UNIT-3

Question-1: Find the mean of the following dataset.

Solution:

We have the following table-

Then Mean will be-

Question-2: Find the arithmetic mean of the following dataset-

Solution:

We have the following distribution-

Question-3: Find the median of the following dataset-

Solution:

So that median class is 20-30.

Now putting the values in the formula-

So that the median is 28.57

Question-4: Find the mode of the following dataset-

Solution:

Here highest frequency is 9. So that the modal class is 40-50,

Put the values in the given data-

Hence the mode is 42.86

Question-5: If coefficient of skewness is 0.64. Standard deviation is 13 and mean is 59.2, then find the mode and median.

Solution:

We know that-

So that-

And we also know that-

Question-6: Calculate the Karl Pearson’s coefficient of skewness of marks obtained by 150 students.

Solution:

Mode is not well defined so that first we calculate mean and median-

Class	f	x	CF		Fd
0-10	10	5	10	-3	-30	90
10-20	40	15	50	-2	-80	160
20-30	20	25	70	-1	-20	20
30-40	0	35	70	0	0	0
40-50	10	45	80	1	10	10
50-60	40	55	120	2	80	160
60-70	16	65	136	3	48	144
70-80	14	75	150	4	56	244

Now,

And

Standard deviation-

Then-

Question-7: A die is thrown 8 times then find the probability that 3 will show-

1. Exactly 2 times

2. At least 7 times

3. At least once

Solution:

As we know that-

Then-

1. Probability of getting 3 exactly 2 times will be-

2. Probability of getting 3 at least 7 or 8 times will be-

3. Probability of getting 3 at least once or (1 or 2 or 3 or 4 or 5 or 6 or 7 or 8 times)-

Question-8: If the percentage of failure in a test is 20. If six students appear in the test, then what will be the probability that at least five students will pass the test?

Solution:

Here

Then the probability of at least five students will pass the test-

Question-9: Find mean and variance of a binomial distribution with p = 1/4 and n = 10.

Solution:

Here

Mean = np =

Variance = npq =

Question-10: : If cars arriving at workshop follow the Poisson distribution. If the average number of cars arrivals during a specified period of an hour is 2.

Find the probabilities that during the given hour-

1. No car arrive

2. At least two cars arrive.

Solution:

Here the average of car arrivals is - 2

So that mean = 2

Let X be the number of cars arriving during the given  hour,

By using Poisson distribution, we get-

So that the required probability-

1. P [no car will arrive] = P [X = x] =

2. P [At least two cars will arrive] = P [X≥2] = P [X =2] + P [X = 3] + ……….

              = 1 - P [[X =1] + P [X =0]]

Question-11: If the probability that a vaccine given to the patients shows bad reaction is 0.001, then find the probability that out of 2000 patients-

1. Exactly 3 patients

2. More than 2 patients

3. No patient

Will show bad reaction.

Solution:

Here p = 0.001 and number of patients (n) = 2000

Then

By using Poisson distribution, we get-

1. Probability that exactly 3 patients show bad reaction is-

2. Probability that more than 2 patients show bad reaction-

3. Probability that no patient shows bad reaction-

Question-12: If a book has 600 pages and it has 40 printing mistakes. Assume that these mistakes are randomly distributed and x the number of mistakes per page follow Poisson distribution.

What is the probability that there will not be any mistake if 10 pages selected at random?

Solution:

Here

We get  by using Poisson distribytion-

Then-

Question-13: 1. If X then find the probability density function of X.

2. If X then find the probability density function of X.

Solution:

1. We are given X

Here

We know that-

Then the p.d.f.  will be-

2. . We are given X

Here

We know that-

Then the p.d.f.  will be-

Question-14: : If a random variable X is normally distributed with mean 80 and standard deviation 5, then find-

1. P[X > 95]

2. P[X < 72]

3. P [85 < X <97]

[Note- use the table- area under the normal curve]

Solution:

The standard normal variate is –

Now-

1. X = 95,

So that-

2. X = 72,

So that-

3. X = 85,

X = 97,

So that-

Question-15: In a company the mean weight of 1000 employees is 60kg and standard deviation is 16kg.

Find the number of employees having their weights-

1. Less than 55kg.

2. More than 70kg.

3. Between 45kg and 65kg.

Solution:

Suppose X be a normal variate = the weight of employees.

Here mean 60kg and S.D. = 16kg

X

Then we know that-

We get from the data,

Now-

1. For X = 55,

So that-

2. For X = 70,

So that-

3. For X = 45,

For X = 65,

Hence the number of employees having weights between 45kg and 65kg-

Question-16: Compute the Spearman’s rank correlation coefficient of the dataset given below-

Solution:

Person	Rank in test-1	Rank in test-2	d =
A	9	1	8	64
B	10	2	8	64
C	6	3	3	9
D	5	4	1	1
E	7	5	2	4
F	2	6	-4	16
G	4	7	-3	9
H	8	8	0	0
I	1	9	-8	64
J	3	10	-7	49
Sum				280

Question-17: Two variables X and Y are given in the dataset below, find the two lines of regression.

Solution:

The two lines of regression can be expressed as-

And

x	y			Xy
65	66	4225	4356	4290
66	68	4356	4624	4488
67	65	4489	4225	4355
67	69	4489	4761	4623
68	74	4624	5476	5032
69	73	4761	5329	5037
70	72	4900	5184	5040
71	70	5041	4900	4970
Sum = 543	557	36885	38855	37835

Now-

And

Standard deviation of x-

Similarly-

Correlation coefficient-

Put these values in regression line equation, we get

Regression line y on x-

Regression line x on y-

Question-18: Find the straight line that best fits of the following data by using method of least square.

Solution:

Suppose the straight line

y = a + bx…….. (1) 

Fits the best-

Then-

x	y	Xy
1	14	14	1
2	27	54	4
3	40	120	9
4	55	220	16
5	68	340	25
Sum = 15	204	748	55

Normal equations are-

Put the values from the table, we get two normal equations-

On solving the above equations, we get-

So that the best fit line will be- (on putting the values of a and b in equation (1))

Question-19: Fit the second degree parabola of the following data by using method of least squares.

Solution:

By taking u = x – 1933 and v = y – 357

Then equation becomes

Putting the values from the table in normal equations-

We get-

11 = 3A + 0B + 60C or 11 = 9A + 60C

51 = 0A + 60B + 0C or B = 17 / 20

-9 = 60A + 0B + 708C or -9 = 60A + 708C

On solving, we get-

On solving the above equation, we get-

Question-20: Fit the curve by using the method of least square.

Solution:

Here-

Now put-

Then we get-

x	Y		XY
1	7.209	1.97533	1.97533	1
2	5.265	1.66108	3.32216	4
3	3.846	1.34703	4.04109	9
4	2.809	1.03283	4.13132	16
5	2.052	0.71881	3.59405	25
6	1.499	0.40480	2.4288	36
Sum = 21		7.13988	19.49275	91

Normal equations are-

Putting the values form the table, we get-

7.13988 = 6c + 21b

19.49275 = 21c + 91b

On solving, we get-

b = -0.3141 and c = 2.28933

c =

Now put these values in equations (1), we get-