undefined | unit 4 basic statistics

Unit-4Basic Statistics Q1) Find the arithmetic mean for the following distribution

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	8	20	10	5

S1) Let assumed mean (a) = 25

Class	Mid-value (x)	Frequency (f)		fd
0-10	5	7	-20	-140
10-20	15	8	-10	-80
20-30	25	20	0	0
30-40	35	10	+ 10	+100
40-50	45	5	+ 20	+100
Total		50		-20

Q2) Find the value of median from the following data

Number of days for which absent (less than)	5	10	15	20	25	30	35	40	45
Number of students	29	224	465	582	634	644	650	653	655

S2) The given cumulative frequency distribution will first be converted into ordinary frequency as under:

Class interval	Cumulative frequency	Ordinary frequency
0-5	29	29=29
5-10	224	224-29=105
10-15	465	465-224=241
15-20	582	582-465=117
20-25	634	634-582=52
25-30	644	644-634=10
30-35	650	650-644=6
35-40	653	653-650=3
40-45	655	655-653=2

Median = size of

327.5th item lies in 10-15 which is the median class

Where l stands for lower limit of median class.N stands for the total frequencyC stands for cumulative frequency just preceding the median classi stands for class intervalf stands for frequency for the median class

Q3) Find the mode from the following data

Age	0-6	6-12	12-18	18-24	24-30	30-36	36-42
Frequency	6	11	25	35	18	12	6

S3)

Age	Frequency	Cumulative frequency
0-6	6	6
6-12	11	17
12-18		42
18-24	35 = f	77
24-30		95
30-36	12	107
36-42	6	113

Q4) If coefficient of skewness is 0.64. Standard deviation is 13 and mean is 59.2, then find the mode and median.S4)We know that-

So that-

And we also know that-

Q5) Calculate the Karl Pearson’s coefficient of skewness of marks obtained by 150 students.

S5) 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-

Q6) If on an average one ship in every ten is wrecked. Find the probability that out of 5 ships expected to arrive, 4 at least we will arrive safely.S6) Out of 10 ships one ship is wrecked.I.e. nine ships out of 10 ships are safe, P (safety) =

P (at least 4 ships out of 5 are safe) = P (4 or 5) = P (4) + P(5)

Q7) The probability that a man aged 60 will live to be 70 is 0.65. what is the probability that out of 10 men, now 60, at least seven will live to be 70?S7) The probability that a man aged 60 will live to be 70

Number of men= n = 10

Probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)

= P (7)+ P(8)+ P(9) + P(10) =

Q8) A die is tossed thrice. A success is getting 1 or 6 on a TOSS. Find the mean and variance of the number of successes.S8)

Q9) Assume that the probability of an individual coal miner being killed in a mine accident during a year is

. Use appropriate statistical distribution to calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year.S9)

Q10) Find the area under the normal curve in each of the cases(a) Z = 0 and z = 1.2(b) Z = -0.68 and z = 0(c) Z = -0.46 and z = -2.21(d) Z = 0.81 and z = 1.94(e) To the left of z = -0.6(f) Right of z = -1.28S10) (a) Area between Z = 0 and z = 1.2 =0.3849(b)Area between z = 0 and z = -0.68 = 0.2518 (c)Required area = (Area between z = 0 and z = 2.21) + (Area between z = 0 and z =-0.46)\= (Area between z = 0 and z = 2.21)+ (Area between z = 0 and z = 0.46)=0.4865 + 0.1772 = 0.6637(d)Required area = (Area between z = 0 and z = 1.+-(Area between z = 0 and z = 0.81) = 0.4738-0.2910=0.1828(e) Required area = 0.5-(Area between z = 0 and z = 0.6) = 0.5-0.2257=0.2743(f)Required area = (Area between z = 0 and z = -1.28)+0.5 = 0.3997+0.5 = 0.8997 Q11) In a normal distribution, 31% of the items are 45 and 8% are over 64. Find the mean and standard deviation of the distribution.S11) Let

be the mean and

the S.D.If x = 45,

If x = 64,

Area between 0 and

[From the table, for the area 0.19, z = 0.496)

Area between z = 0 and z =

(from the table for area 0.42, z = 1.405)

Solving (1) and (2) we get

Q12) Psychological test of the intelligence and of Engineering ability were applied to 10 students. Here is a record of ungrouped data showing intelligence ratio (I.R) and Engineering ratio (E.R). Calculate the coefficient of correlation.

Student	A	B	C	D	E	F	G	H	I	J
I.R.	105	104	102	101	100	99	98	96	93	92
E.R.	101	103	100	98	95	96	104	92	97	94

S12) We construct the following table

Student	Intelligence ratio x	Engineering ratio y y			XY
A	105 6	101 3	36	9	18
B	104 5	103 5	25	25	25
C	102 3	100 2	9	4	6
D	101 2	98 0	4	0	0
E	100 1	95 -3	1	9	-3
F	99 0	96 - 2	0	4	0
G	98 -1	104 6	1	36	-6
H	96 -3	92 -6	9	36	18
I	93 -6	97 -1	36	1	6
J	92 -7	94 -4	49	16	28
Total	990 0	980 0	170	140	92

From this table, mean of x, i.e.

and mean of y, i.e.

Substituting these value in the formula (1)p.744 we have

Q13) The correlation table given below shows that the ages of husband and wife of 53 married couples living together on the census night of 1991. Calculate the coefficient of correlation between the age of the husband and that of the wife.

Age of husband	Age of wife							Total
Age of husband	15-25	25-35		35-45	45-55	55-65	65-75	Total
15-25	1		1	-	-	-	-	2
25-35	2		12	1	-	-	-	15
35-45	-		4	10	1	-	-	15
45-55	-		-	3	6	1	-	10
55-65	-		-	-	2	4	2	8
65-75	-		-	-	-	1	2	3
Total	3		17	14	9	6	4	53

S13)

Age of husband				Age of wife x series							Suppose
Age of husband				15-25	25-35	35-45	45-55	55-65	65-75	Total f
Years			Midpoint x	20	30	40	50	60	70
Age group	Midpoint y			-20	-10	0	10	20	30
Age group	Midpoint y			-2	-1	0	1	2	3
15-25	20	-20	-2	4 1	2 1					2	-4	8	6
25-35	30	-10	-1	4 2	12 12	0 1				15	-15	15	16
35-45	40	0	0		0 4	0 10	0 1			15	0	0	0
45-55	50					0 3	6 6	2 1		10	10	10	8
55-65	60						4 2	16 4	12 2	8	16	32	32
65-75	70							6 1	18 2	3	9	27	24
Total f				3	17	14	9	6	4	53 = n	16	92	86
				-6	-17	0	9	12	12	10	Thick figures in small sqs. for Check: From both sides
				12	17	0	9	24	36	98
				8	14	0	10	24	30	86

With the help of the above correlation table, we have

Q14) While calculating correlation coefficient between two variables x and y from 25 pairs of observations, the following results were obtained : n = 25,

Later it was discovered at the time of checking that the pairs of values x -8,6 and y = 12, 8 were copied down as x = 6,8 and y = 14,6. Obtain the correct value of correlation coefficients.S14) To get the correct results, we subtract the incorrect values and add the corresponding correct values.The correct results would be

Q15) Three judges A,B,C give the following ranks. Find which pair of judges has common approach

A	1	6	5	10	3	2	4	9	7	8
B	3	5	8	4	7	10	2	1	6	9
C	6	4	9	8	1	2	3	10	5	7

S15) Here n = 10

A (=x)	Ranks by B(=y)	C (=z)	x-y	y - z	z-x
1	3	6	-2	-3	5	4	9	25
6	5	4	1	1	-2	1	1	4
5	8	9	-3	-1	4	9	1	16
10	4	8	6	-4	-2	36	16	4
3	7	1	-4	6	-2	16	36	4
2	10	2	-8	8	0	64	64	0
4	2	3	2	-1	-1	4	1	1
9	1	10	8	-9	1	64	81	1
7	6	5	1	1	-2	1	1	4
8	9	7	-1	2	-1	1	4	1
Total			0	0	0	200	214	60

Since

is maximum, the pair of judge A and C have the nearest common approach.

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