Statistics And Probability

We know that, d_i = x_i - a

Where,

_xi are data and “a” is the assumed mean

i.e. d_i are the deviations from of mid - points of the classes.

In grouping the data all the observations between lower and upper limits of class marks are taken in one group then mid value or class mark is taken for further calculation.

In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.

We know that the mean

- - - (2)

From (1) and (2) we get

Given,

Above formula is a step deviation (shortcut) formula.

Where x_i is data values, a is assumed mean and h is class size, when class size is same we simplify the calculations of the mean by computing the coded mean of u₁,u₂,u₃…..where

Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa.

Here,

Now, N/2 = 66/2 = 33, which lies in the interval 10 - 15. Therefore, lower limit of the median class is 10.

The highest frequency is 20, which lies in the interval 15 - 20. Therefore, lower limit of modal class is 15. Hence, required sum is 10 + 15 = 25.

Given, classes are not continuous, so we make continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

Here, N/2 = 57/2 = 28.5

which lies in the interval 11.5 - 17.5.

Hence, the upper limit is 17.5.

Here, we see that the highest frequency is 30, which lies in the interval 30 - 40.

Here, N/2 = 67/2 = 33.5 which lies in the interval 125 – 145.

Hence, upper limit of median class is 145.

Here, we see that the highest frequency is 20 which lies in 125 - 145. Hence, the lower limit of modal class is 125.

∴ Required difference = Upper limit of median class – Lower limit of modal class = 145 – 125 = 20

The number of athletes who completed the race in less than 14.6

_{2 + 4 + 5 + 71 = 82}

Hence, frequency in the class interval 30 - 40 is 3.

The event which cannot occur is said to be impossible event and probability of impossible event is zero.

Since, probability of an event always lies between 0 and 1.

Probability of any event cannot be more than 1 or negative as

The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.

Since, probability of an event + probability of its complementary event = 1

So, probability of its complementary event = (1 – Probability of an event) = 1 – P.

We know that the probability expressed as a percentage always lie between 0 and 100. So, it cannot be less than 0.

Since, probability of an event always lies between 0 and 1.

In a deck of 52 cards, there are 12 face cards i.e. 6 red (3 hearts and 3 diamonds) and 6 black cards (3 spade and 3 clubs)

So, probability of getting a red face card = 6/52 = 3/26

A non - leap year has 365 days and therefore 52 weeks and 1 day. This 1 day may be Sunday or Monday or Tuesday or Wednesday or Thursday or Friday or Saturday.

Thus, out of 7 possibilities, 1 favorable event is the event that the one day is Sunday.

∴ Required probability = 1/7

When a die is thrown, then total number of outcomes = 6

Odd number less than 3 is 1 only.

Number of possible outcomes = 1

∴ Required probability = 1/6

In a deck of 52 cards, there are 13 cards of heart and 1 is ace of heart.

Hence, the number of outcomes favorable to _{E = 52 - 1 = 51.}

Here, total number of eggs = 400

Probability of getting a bad egg = 0.035

∴ Number of bad eggs = 0.035 × 400 = 14

Given, total number of sold tickets = 6000

Let she bought x tickets.

Then, probability of her winning the first prize is given as,

[given]

∴ x = 480

Hence, she bought 480 tickets.

Number of total outcomes = 40

Multiples of 5 between 1 to 40 = 5, 10, 15, 20, 25, 30, 35, 40

∴ Total number of possible outcomes = 8

∴ Required probability = 8/40 = 1/5

Total numbers of outcomes = 100

So, the prime numbers between 1 to 100are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 56, 61, 67, 71, 73, 79, 83, 89 and 97.

∴ Total number of possible outcomes = 25

∴ Required probability = 13/50

Total number of students = 23

Number of students in house A, B and C = 4 + 8 + 5 = 17

∴ Remains students = 23 - 17 = 6

So, probability that the selected student is not from A, B and C = 6/23

Not always, because for calculating median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed (or equally spaced).

No, it is not necessary that assumed mean consider as the mid - point of the class interval. It is considered as any value which is easy to simplify it.

No, the value of these three measures can be the same, it depends on the type of data.

Not always, it depends on the given data.

No, the probability of each is not 1/4 because the probability of no girl in three children is zero and probability of three girls in three children is one.

Justification

So, these events are not equally likely as outcome one girl, means gbb, bgb, bbg - three girls’ means “ggg” and so on.

No, the outcomes are not equally likely, because 3 contains half part of the total region, so it is more likely than 1 and 2, since 1 and 2, each contains half part of the remaining part of the region.

Apoorv throw two dice one.

So total number of outcomes = 36

Number of outcomes for getting product 36 = 1(6×6)

∴ Probability for Apoorv = 1/36

Also, Peehu throws one die,

So, total number of outcomes = 6

Number of outcomes for getting square = 36

∴ Probability for Peehu = 6/36 = 1/6

Hence, Peehu has better chance of getting the number 36.

Yes, probability of each outcome is because head and tail both are equally likely events.

No, this is not correct.

Suppose we throw a die, then total number of outcomes = 6

Possible outcomes = 1or 2or 3or 4or 5or 6

∴ Probability of getting 1 = 1/6

Now, probability of getting not 1 = 1 – Probability of getting 1

= 1 - 1/6 = 5/6

I toss three coins together [given]

So, total number of outcomes = 2³ = 8

and possible outcomes are(HHH),(HTT),(THT),(TTH),(HHT),(THH),(HTH) and (TTT).

Now, probability of getting no head = 1/8

Hence, the given conclusion is wrong because the probability of no head is 1/8 and 1/4.

No, if let we toss a coin, then we get head or tail, both are equally likely events. So, probability is 1/4 if we toss a coin 6 times, then probability will be same in each case. So, the probability of getting a head is not 1.

The outcome of next toss may or may not be tail, because on tossing a coin, we get head or tail so both are equally likely events.

No, let we toss a coin, then we get head or tail, both are equally likely events. i.e., probability of each event is 1/2 .So, no question of expecting a tail to have a higher change in 4th toss.

We know that, between 1 to 100 half numbers are even and half numbers are odd i.e., 50 numbers (2, 4, 6, 8, …, 96, 98, 100) are even and 50 numbers (1, 3, 5, 7, …, 97, 99) are odd. So, both events are equally likely.

So, probability of getting even number = 50/100 = 1/2

and probability of getting odd number = 50/100 = 1/2

Hence, the probability of each is 1/2

We first, find the class mark x_i of each class and then proceed as follows.

Therefore, mean

Hence, mean of the given distribution is 5.5.

We first, find the class mark x_i of each class and then proceed as follows

Therefore, mean

Hence, the mean of scores of 20 students in mathematics test 35.

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Now, we first find the class mark x_iof each class and then proceed as follows

Therefore, mean

Hence, mean of the given data is 12.93.

The table is given below:

Since, given data is not continuous, so we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Hence, the mean of pages written per day is 26.

No need to convert discontinuous classes into continuous for class mark because class mark of C.I are same and gives same result

∴ Assumed mean, a = 300.5

and d_i = (x_i - a)

= 300.5 + 2800/50

= 356.5

Hence, the average daily income of employees is Rs.356.5

We first, find the class mark of each class and then proceed as follows.

∴ Assumed mean, a = 110

Class width, h = 4

and total observations, N = 100

By step deviation method,

=

But seats cannot be in decimal, so number of seats is 109.

We first find the class mark x_i, of each class and then proceed as follows

∴ Assumed mean, (a) = 125

Class width, (h) = 10

and total observations, (N) = 50

By step deviation method,

= 125 - 16

= 123.4kg

Hence, mean weight of wrestlers = 123.4kg

Here,

and

Hence, mean mileage is 14.48 km/h

No, the manufacturer is claiming mileage 1.52 km/h more than average mileage.

The cumulative distribution (less than type) table is shown below

Here, we observe that 10 students have scored marks below 10 i.e., it lies between class interval 0 - 10. Similarly, 50 students have scored marks below 20. So, 50 – 10 = 40 students lie in the interval 10 – 20 and so on. The table of a frequency distribution for the given data is:

Here, we observe that, all 34 students have scored marks more than or equal to 0. Since, 32 students have scored marks more than or equal to 10. So, 34 – 32 = 2 students lies in the interval 0 - 10 and so on.

Now, we construct the frequency distribution table.

On comparing last two tables, we get

a = 12

∴ 12 + b = 25

⇒ b = 25 - 12 = 13

22 + b = c

⇒ c = 22 + 13 = 35

22 + b + d = 43

⇒ 22 + 13 + d = 43

⇒ d = 43 - 35 = 8

and 22 + b + d + e = 48

⇒ 22 + 13 + 8 + e = 48

⇒ e = 48 - 43 = 5

and 24 + b + d + e = f

⇒ 24 + 13 + 8 + 5 = f

∴ f = 50

(i) We observe that the number of patients which take medical treatment in a hospital on a particular day less than 10 is 0. Similarly, less than 20 include the number of patients which take medical treatment from 0 - 10 as well as the number of patients which take medical treatment from 10 - 20.

So, the total number of patients less than 20 is 0 + 60 = 60, we say that the cumulative frequency of the class 10 - 20 is 60. Similarly, for other class.

(ii) Also, we observe that all 300 patients which take medical treatment more than or equal to 10. Since, there are 60 patients which take medical treatment in the interval 10 - 20, this means that there are 300 – 60 = 240 patients which take medical treatment more than or equal to 20. Continuing in the same manner.

Here, we observe that, 17 students have scored marks below 20 i.e., it lies between class interval 0 - 20 and 22 students have scored marks below 40, so 22 – 17 = 5 students lies in the class interval 20 - 40 continuing in the same manner, we get the complete frequency distribution table for given data.

First we construct a cumulative frequency table.

It is given that, _{n = 600}

∴ n/2 = 600/2 = 300

Since, cumulative frequency 440 lies in the interval 1000 – 2000.

Here, (lower median class) l = 1000

f = 190, c_f = 250, (class width) h = 1000

and (total observation) n = 600

= 1000 +

= 1000 + 5000/19

= 1000 + 263.15 = 1263.15

Hence, the median income is Rs.1263.15.

First we construct the cumulative frequency table

It is given that, _{n = 33}

∴ n/2 = 33/2 = 16.5

So, the median class is 100 - 115.

Where, lower limit(l) = 100

Frequency(f) = 9

Cumulative frequency(cf) = 11

and class width(h) = 15

= 100 + 82.5/9

= 100 + 9.17

= 109.17

Hence, the median bowling speed is 109.17 km/h.

In a given data, the highest frequency is 41, which lies in the interval 10000 - 15000.

Here, l = 10000,f_m = 41,f₁ = 26,f₂ = 16 and h = 5000

= 10000 + 15×125

= 10000 + 1875

= 11875

Hence, the modal income is Rs.11875 per month.

In the given data, the highest frequency is 26, which lies in the interval 201 - 202

Here, l = 201,f_m = 26,f₁ = 12,f₂ = 20 and (class width) h = 1

= 201 + ()

= 201 + ()

= 201 + 14/20

= 201 + 0.7

= 201.7g

Hence, the modal weight is 201.7 g.

Two dice are thrown at the same time.

[given]

So, total number of possible outcomes = 36

(i) We have, same number on both dice.

So, possible outcomes are (1,1), (2,2), (3, 3), (4, 4), (5, 5) and (6, 6).

∴ Number of possible outcomes = 6

Now, required probability = 6/36 = 1/6

(ii) We have different number on both dice.

So, number of possible outcomes

= 36 – Number of possible outcomes for same number on both dice

_{= 36 - 6 = 30}

∴ Required probability = 30/36 = 5/6

Two dice are thrown simultaneously.

[given]

So, that number of possible outcomes = 36

(i) Sum of the numbers appearing on the dice is 7.

So, the possible ways are (1, 6), (2,5), (3, 4), (4, 3), (5, 2) and (6, 1).

Number of possible ways = 6

∴ Required probability = 6/36 = 1/6

(ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

So, the possible ways are (1, 1), (1,2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) and (6, 5).

Number of possible ways = 15

∴ Required probability = 15/36 = 5/12

(iii) Sum of the numbers appearing on the dice is 1.

It is not possible, so its probability is zero.

Number of total outcomes = 36

(i) When product of the numbers on the top of the dice is 6.

So, the possible ways are (1, 6), (2,3), (3, 2) and (6, 1).

Number of possible ways = 4

∴ Required probability = 4/36 = 1/9

(ii) When product of the numbers on the top of the dice is 12.

So, the possible ways are (2, 6), (3,4), (4, 3) and (6, 2).

Number of possible ways = 4

∴ Required probability = 4/36 = 1/9

(iii) Product of the numbers on the top of the dice cannot be 7. So, its probability is zero.

Number of total outcomes = 36

When product of numbers appearing on them is less than 9, then possible ways are (1, 6), (1, 5), (1, 4), (1, 3), (1, 2), (1, 1), (2, 2), (2, 3), (2, 4), (3, 2), (4, 2), (4, 1), (3, 1), (5, 1), (6, 1) and (2, 1).

Number of possible ways = 16

∴ Required probability = 16/36
= 4/9

Number of total outcome = n(S) = 36

(i) Let E₁ = Event of getting sum 2 = {(1,1),(1,1)}

n(E₁) = 2

(ii) Let E₂ = Event of getting sum 3 = {(1,2),(1,2),(2,1),(2,1)}

n(E₂) = 4

(iii) Let E₃ = Event of getting sum 4 = {(2,2)(2,2),(3,1),(3,1),(1,3),(1,3)}

n(E₃) = 6

(iv) Let E₄ = Event of getting sum 5 = {(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}

n(E₄) = 6

(v) Let E₅ = Event of getting sum 6 = {(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}

n(E₅) = 6

(vi) Let E₆ = Event of getting sum 7 = {(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}

n(E₆) = 6

(vii)Let E₇ = Event of getting sum 8 = {(5,3),(5,3),(6,2),(6,2)}

n(E₇) = 4

(viii)Let E₈ = Event of getting sum 9 = {(6,3),(6,3)}

n(E₃) = 2

The possible outcomes, if a coin is tossed 2 times is

S = {(HH),(TT),(HT),(TH)}

n(S) = 4

Let E = Event of getting at most one head = {(TT),(HT),(TH)}

n(S) = 3

Hence, required probability

The possible outcomes if a coin is tossed 3 times is

S = {(HHH),(TTT),(HTT),(THT),(TTH),(THH),(HTH),(HHT)}

n(S) = 8

(i) Let E₁ = Event of getting all heads = {(HHH)}

∴ n(E₁) = 1

(ii) Let E₂ = Event of getting atleast 2 heads = {(HHT),(HTH),(THH),(HHH)}

∴ n(E₂) = 4

The total number of sample space in two dice, n(S) = 6

Let

E = Event of getting the numbers whose difference is 2

= {(1,3),(2,4),(3,5),(4,6),(3,1),(4,2),(5,3),(6,4)}

∴ n(E) = 8

No. of red ball = 10

No. of blue ball = 5

No. of green balls = 7

If a ball is drawn out of 22 balls (5 blue
+ 7 green + 10 red), then the total number of outcomes are

n(S) = 22

(i) Let E₁ = Event of getting a red ball

∴ n(E₁) = 10

(ii) Let E₂ = Event of getting a green ball

n(E₂) = 7

(iii) Let E₃ = Event getting a red ball or a green ball i.e., not a blue ball.

n(E₃) = (10 + 7) = 17

If we remove one king, one queen and one jack of clubs from 52 cards, then the remaining cards left, n(S) = 49

(i) Let E₁ = Event of getting a heart

n(E₁) = 13

(ii) Let E₂ = Event of getting a king

n(E₂) = 3

[since, out of 4 king, one club cards is already removed]

(i)Total number of cards = 52 - 3 = 49

Let E₃ = Event of getting a club

n(E₃) = (13 - 3) = 10

∴ Required probability

(ii) Let E₄ = Event of getting 10 of hearts

n(E₄) = 1

[because in 52 playing cards only 13 are the heart cards and only one 10 in 13 heart cards]

∴ Required probability =

In out of 52 playing cards, 4 jacks, 4 queens and 4 kings are removed, then the remaining cards are left,

n(S) = 52 - 3×4 = 40

(i) Let E₁ = Event of getting a card whose value is 7

E = Card value 7 may be of a spade, a diamond, a club or a heart

∴ n(E₁) = 4

(ii) Let E₂ = Event of getting a card whose value is greater than 7

= Event of getting a card whose value is 8, 9 or 10

(E₂)

(iii) Let E₃ = Event of getting a card whose value is less than 7

= Event of getting a card whose value is 1, 2, 3, 4, 5 or 6

∴ n(E₃) = 6×4 = 24

The number of integers between 0 and 100 is n(S) = 99

(i) Let E₁ = Event of choosing an integer which is divisible by 7

= Event of choosing and integer which is multiple of 7

= {7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98}

∴ n(E₁) = 14

(ii) Let Event of choosing an integer which is not divisible by 7

∴ n(E₂) = n(S) - n(E₁)

= 99 - 14 = 85

Total number of out comes with numbers 2 to 101, n(s) = 100

(i) Let E₁ = Event of selecting a card which is an even number = {2, 4, 6, …100}

[In an AP, l= a + (n - 1)d,

here l = 100,a = 2and d = 2

⇒ 100 = 2 + (n - 1)2

⇒ (n - 1) = 49

⇒ n = 50

∴ n (E₂) = 9

∴ Required probability

(ii) Let E₂ = Event of selecting a card which is a square number

= {4,9,16,25,36,49,64,81,100}

= {(2)²,(3)²,(4)²,(5)²,(6)²,(7)²,(8)²,(9)²,(10)²}

Hence, required probability =

We know that, in English alphabets, there are (5 vowels + 21 consonants) = 26 letters. So, total number of outcomes in English alphabets is,n(S) = 26

Let E = Event of choosing an English alphabet, which is a consonant

= {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s t, v, w, x, y, z}

∴ n(E) = 21

Hence, required probability

Total number of sealed envelopes in a box, n(S) = 1000

Number of envelopes containing cash prize = 10 + 100 + 200 = 310

Number of envelopes containing no cash prize,

n(E) = 1000 - 310 = 690

P(E) =

⇒ P(E) = 0.69

Total number of slips in a box, n(S) = 25 + 50 = 75

From the chart it is clear that, there are 11 slips which are marked other than Rs1.

∴ Required probability

Total number of bulbs, n(s) = 24

Let E₁ = Event of selecting not defective bulb = Event of selecting good bulbs

n(E₁) = 18

Suppose, the selected bulb is defective and not replaced, then total number of bulbs remains in a carton, n(S) = 23

In them, 18 are good bulbs and 5 are defective bulbs.

∴ P(selecting second defective bulb) = 5/23

Total number of figures

n(S) = 8 triangles + 10 squares = 18

(i) P(lost piece is a triangle) = 8/18 = 4/9

(ii) P(lost piece is a square) = 10/18 = 5/9

(iii) P(square of blue colour) = 6/18 = 1/3

(iv) P(triangle of red colour) = 5/18

Total possible outcomes of tossing a coin 3 times,

S = {(HHH),(TTT),(HTT),(THT),(TTH),(THH),(HTH),(HHT)}

∴ n(S) = 8

(i) Let E₁ = Event that Sweta losses the entry fee

= She tosses tail on three times

n(E₁) = {(T T T)}

(ii) Let Event that Sweta gets double entry fee

= She tosses tail on three times = {(HHH)}

n(E₂) = 1

(iii) Let Event that Sweta gets her entry fee back

= Sweta gets heads one or two times

= {(HTT),(THT),(TTH),(HHT),(HTH),(THH)}

n(E₃) = 6

Given, a die has its six faces marked {0, 1, 1, 1, 6, 6}

∴ Total sample space, n(S) = 6² = 36

(i)Number of favorable outcomes are (0,0),(0,1),(0,6),(1,0),(1,1),(1,6) ,(6,0),(6,1),(6,6).The different score which are possible are 6 scores i.e., 0,1,2,6,7 and 12.

(ii) Let E = Event of getting a sum 7

= {(1,6),(1,6),(1,6),(1,6),(1,6),(1,6),(6,1),(6,1),(6,1),(6,1)(6,1)(6,1)}

∴ n(E) = 12

Given, total number of mobile phones

_{n(S) = 48}

(i) Let E₁ = Event that Varnika will buy a mobile phone

= Varnika buy only, if it is good mobile

∴ n(E₁) = 42

(ii) Let E₂ = Event that trader will buy only when it has no major defects

= Trader will buy only 45 mobiles

∴ n(E) = 45

Given that, A bag contains total number of balls = 24

A bag contains number of red balls = x

A bag contains number of white balls _{= 2x}

and a bag contains number of blue balls _{= 3x}

By condition, _{x + 2x + 3x = 24}

⇒ 6x = 24

⇒ x = 4

∴ Number of red balls = x = 4

Number of white balls = 2x = 2×4 = 8

and number of blue balls = 3x = 3×4 = 12

So, total number of outcomes for a ball is selected at random in a bag contains 24 balls.
⇒ n(S) = 24

(i) Let E₁ = Event of selecting a ball which is not red i.e., can be white or blue.

∴ n(E₁) =Number of white balls + Number of blue balls

⇒ ∴ n(E₁) = 8 + 12 = 20

(ii) Let E₂ Event of selecting a ball which is white

Number of white ball = 8

So, required probability

Given that, at a fete, cards bearing numbers 1 to 1000 one number on one card, are put in a box. Each player selects one card at random and that card is not replaced so, the total number of outcomes are n(S) = 1000

If the selected card has a perfect square greater than 500, then player wins a prize.

(i) Let E₁ = Event first player wins a prize = Player select a card which is a perfect square greater than 500

= {529, 576, 625, 729, 784, 841, 900, 961}

= {(23)²,(24)²,(25)²,(26)²,(27)²,(28)²,(29)²,(30)²,(31)²}

∴ n(E) = 9

So, required probability

(ii) First, has won i.e., one card is already selected, greater than 500, has a perfect square. Since, repetition is not allowed. So, one card is removed out of 1000 cards. So, number of remaining card is 999.

∴ Total number of remaining outcomes, n(S’) = 999

Let E₂ be the event that the second player wins a prize, if the first has won.

Then, the remaining cards has a perfect square greater than 500 = 8

So, required probability

Here, we observe that, 5 students have scored marks below 10, i.e. it lies between class Interval 0 - 10 and 9 students have scored marks below 20.

So, (9 – 5) = 4 students lie in the class interval 10 - 20. Continuing in the same manner, we get the complete frequency distribution table for given data.

Here, (assumed mean) , a = 45

and (class width) h = 10

By step deviation method,

Here, we observe that, all 100 residents of a town have age equal and above 0. Since, 90 residents of a town have age equal and above 10.

So, 100 – 90 = 10 residents lie in the interval 0 - 10 and so on. Continue in this manner, we get frequency of all class intervals. Now, we construct the frequency distribution table.

Here, (assumed mean) a = 35

and (class width) h = 10

By step deviation method,

= 35 - 4 = 31

Hence, the required mean age is 31 yr.

First, we find the class marks of the given data as follows.

Here, (assume mean)

and (class width)

By assumed mean method,

= 203.5 - (108/70)

= 203.5 - 1.54 = 201.96

Hence, the required mean weight is 201.96 g.

We observe that, the number of packets less than 200 is 0. Similarly, less than 201 include the number of packets from 0 - 200 as well as the number of packets from 200 - 201.

So, the total number of packets less than 201 is 0 + 13 = 13. We say that, the cumulative frequency of the class 200 - 201 is 13. Similarly, for other class.

To draw the less than type ogive, we plot the points (200, 0), (201, 13), (202, 40) (203, 58), (204, 68), (205, 69) and (206, 70) on the paper and join by free hand.

∴ Total number of packets (n) = 70

Now, N/2 = 35

Firstly, we plot a point (0, 35) on Y - axis and draw a line y = 35 parallel to X - axis. The line cuts the less than ogive curve at a point. We draw a line on that point which is perpendicular to X - axis. The foot of the line perpendicular to X - axis is the required median.

∴ Median weight = 201.8 g

For less than type table we follow the Q.5.

Here, we observe that, the weight of all 70 packets is more than or equal to 200. Since, 13 packets lie in the interval 2001 - 201. So, the weight of 70 – 13 = 57 packets is more than or equal to 201. Continuing in this manner we will get remaining more than or equal to 202, 203, 204, 205 and 206.

To draw the less than type ogive, we plot the points (200, 0), (201, 13), (202, 40), (203, 58), (204, 68), (205, 69), (206, 70) on the paper and join them by free hand.

To draw the more than type ogive plot the point (200, 70), (201, 57), (202, 30), (203, 12), (204, 2), (205, 1), (206, 0) on the the graph paper and join them by free hand.

Hence, required median weight = Intersection point of X - axis = 201.8 g.

First, we construct a cumulative frequency table

∴ N/2 = 280/2 = 140

(i) Here, median class is 10 - 15, because 140 lies in it.

Lower limit(l) = 10,

Frequency(f) = 133

Cumulative frequency (f_c) = 49 and class width (h) = 5

= 10 + 3.421

= Rs.13.421 (in thousand)

= 13.421× 1000

= Rs.13421

(ii) Here, the highest frequency is 133, which lies in the interval 10 - 15, called modal class.

Lower limit(l) = 10, class width(h) = 5,

f_m = 133,

f₁ = 49 and

f₂ = 63

= 10 + 2.727

= Rs.12.727 (in thousand)

= 12.727 × 1000 = Rs.12727

Hence, the median and modal salary are Rs.13421 and Rs12727, respectively.

First, we calculate the class mark of given data

Given that, sum of all frequencies = 120

Here,

assumed mean, a = 50 and

class width, h = 20

By step deviation method,

⇒ 4 + f₂ - f₁ = 0

⇒ -f₂ + f₁ = 4 - - - - (ii)

On adding Eqs. (i) and (ii), we get

2f₁ = 56

⇒ f₁ = 28

Put the value of in Eq. (i), we get

f₂ = 52 - 28

⇒ f₂ = 24

Hence, f₁ = 28 and f₂ = 24

Given, _{N = 90}

N/2 = 90/2 = 45

Which lies in the interval 50 - 60.

Lower limit, l = 50,

f = 20,

f_c = 40 + p,

h = 10

∴ p = 5

Also, _{78 + p + q = 90} [given]

⇒ 78 + 5 + q = 90

⇒ q = 90-83

∴ q = 7

To draw the less than type ogive, we plot the points (124,0), (128,5), (132,13), (136,30), (140,54), (144, 70), (148, 82), (152, 92), (160, 95), (164, 96), (164, 96) and join all these point by free hand.

Here, N/2 = 96/2 = 48

We take, y = 48 in y - coordinate and draw a line parallel to X - axis, meets the curve at A and draw a perpendicular line from point A to the X - axis and this line meets the X - axis at the point which is the median _{= 141.17}

(i) Here, _{N = 200}

Now,

N/2 = 200/2 = 100

(which lies in the interval 15 – 20)

Lower limit, l = 15,

h = 5,

f = 80 and

f_c = 55

= 15 + 2.81

= 17.81 hec

∴ the median is 17.81 hec

(ii) In a given table 80 is the highest frequency.

So, the modal class is 15 - 20.

Here, l = 15,

f_m = 80,

f₁ = 30,

f₂ = 40

h = 5

= 15 + 25/9

= 15 + 2.77 = 17.77 hec

∴ the mode is 17.77 hectare

We observe that, the annual rainfall record of a city less than 0 is 0. Similarly, less than 10 include the annual rainfall record of a city from 0 as well as the annual rainfall record of a city from 0 - 10.

So, the total annual rainfall record of a city for less than 10 cm is 0 + 22 = 22 days. Continuing in this manner, we will get remaining less than 20, 30, 40, 50, and 60.

Also, we observe that annual rainfall record of a city for 66 days is more than or equal to 0 cm. since, 22 days lies in the interval 0 - 10. So, annual rainfall record for 66 – 22 = 44 days is more than or equal to 10 cm. Continuing in this manner we will get remaining more than or equal to 20, 30, 40, 50 and 60.

Now, we construct a table for less than and more than type.

To draw less than type ogive we plot the points (0, 0), (10, 22), (20, 32), (30, 40), (40, 55), (50, 60), (60, 66) on the paper and join them by free hand.

To draw the more than type ogive we plot the points (0, 66), (10, 44), (20, 34), (30, 26), (40, 11), (50, 6) and (60, 0) on the graph paper and join them by free hand.

∴ Total number of days (n) = 66

Now, n/2 = 66/2 = 33

Firstly, we plot a line parallel to X - axis at intersection point of both ogives, which further intersect at (0, 33) on Y - axis. Now we draw a line perpendicular to X - axis at intersection point of both ogives, which further intersect at (21.25, 0) on X - axis. Which is the required median using ogives.

Hence, median rainfall = 21.25 cm

First, we calculate class marks as follows

Here, (assumed mean) a = 170

and (class width) h = 30

By step deviation method,

Average

= 170 + 0.3 = 170.3

Hence, average duration is 170.3s.

For calculating median from a cumulative frequency curve

We prepare less than type or more than type ogive

We observe that, number of calls in less than 95 s is 0. Similarly, in less than 125 s include the number of calls in less than 95 s as well as the number of calls from 95 - 125 s So, the total number of calls than 125 s is 0 + 14 = 14. Continuing in this manner, we will get remaining in less than 155, 185, 215 and 245 s.

Now, we construct a table for less than ogive (cumulative frequency curve).

To draw less than ogive we plot them the points (95, 0), (125, 14) (155, 36), (185, 64), (215, 85), (245, 100) on the paper and join them by free hand.

∴ Total number of calls (n) = 100

∴ n/2 = 100/2 = 50

Now, point 50 taking on Y - axis draw a line parallel to X - axis meet at a point P and draw a perpendicular line from P to the X - axis, the intersection point of X - axis is the median.

Hence, required median is 170.

(i)

(ii)

To draw less than type ogive, we plot the points (0, 0), (20, 6), (40, 17), (60, 34), (80, 46), (100, 50), join all these points by free hand.

Now, N/2 = 50/2 = 25

Taking Y = 25 on Y - axis and draw a line parallel to X - axis, which meets the curve at point A. From point A, we draw a line perpendicular to X - axis, where this meets that point is the required median i.e., 49.4.

(iii) Now, N/2 = 50/2 = 25

Which lies is the interval 40 - 60.

= 40 + 9.41

= 49.41

(iv) Yes, median distance calculated by parts (ii) and (iii) are same.