Statistics

 

Statistics

M. Velrajan

1. Find the range of the following data:25,67,78,43,21,17,49,54,76,92,20,45,86,37,35

      (a)78                             (b)75                   (c)92                   (d)86

Range = largest value - smallest value 

           = 92 - 17 = 75


2. The average of 4 values is 20 and when a quantity is added to each value the average is 22. 

Find the quantity.                    (a)1              (b)2                 (c)3                     (d)4

Adding (or subtracting) a constant to each value increases (or decreases) the mean by the same

constant.

Hence the added value = 22 - 20 = 2


3. Mean of 25 observations was found to be 78.4. But later it was found that 96 was mis – read as 69.

Then the corrected mean is     (a)79.48           (b)76. 54          (c)81.32        (d)78.4

Increase in the total due to the correction = 96 - 69 = 27

Increase in Mean = 27 / 25 = 1.08 (not necessary to find out the exact value, approximately 1 is

enough to get the answer)

Corrected mean = 78.4 + 1.08 = 79.48


4. If the mean of  x, x+2, x+4, x+6, x+8  is 20 then find the value of x.   (a)32    (b)16     (c)8      (d)4

In the given values successive values are increased by a constant.

And there are 5 values and 5 is an odd number.

Hence the mean = the middle value = x + 4.

So, x + 4 = 20.

Hence x = 16.

If an even number of values are given and the successive values are increased by a constant then

mean is the average of the middle two values.

Aliter :  Mean = total / 5 = (5x + 20) / 5 = 20.   

x = 16


5. During the period of decline in Stock  Market prices, a stock sold at Rs. 50 per share on the first day,

Rs. 40 on the second day and Rs. 25 on the third day. If  an investor bought 100, 120 and 180 shares on

the respective three days, the average price paid per share will be

(a)Rs. 35.75    (b)Rs. 38.33        (c)Rs. 34     (d)Rs. 28

Average price = total share price / total number of shares = (50×100 + 40×120 + 25×180) / 400 

                                                                                                  = (5000 + 4800 + 4500) / 400 = 14300 / 400 = 35.75


6. The harmonic mean and geometric mean of two values are 6.4 and 8 respectively, The values are

(a)8 and 8                   (b)32 and 2               (c)4 and 16                           (d)10 and 6

AM × HM = GM2 

AM × 6.4 = 82 = 64

(a + b) / 2 = 64 / 6.4 = 10

a + b = 20

Since in the answer choices only for 4 and 16, 4 + 16 = 20, it is not necessary to find a - b, 

using ab = GM2 = 64 and (a - b)2 = (a + b)2 - 4ab, and then solving to get a and b.       Ans. (c)


7. From the group of 5 men and 5 women, two persons are chosen at random. The probability that one

of them is man and the other women is      (a)2/5    (b)3/5          (c)5/9              (d)4/9

Since there are 5 men and 5 women, number of ways of choosing 

one man and one woman = 5 × 5

Total number of persons = 10

Number of ways of choosing 2 persons = 10C2 = (10 × 9) / 2 = 5 × 9

Required probability = (5 × 5) / (5 × 9) = 5 / 9   Ans. (c)


8. A box contains 4 red, 5 blue and 6 green balls of identical size. Two balls are drawn at random. 

The probability that one is blue and the other is green is 

(a) (5 × 6) / (15 × 14) (b)(5 × 6 × 2) / (15 × 14)  (c) (4 × 5) / (15 × 14) (d)  (4 × 5 × 2) / (14 × 15)

Since there are 5 blue balls and 6 green balls, number of ways of drawing one blue and 

one green balls = 5 × 6

Total number of balls in the box = 15

Number of ways of drawing 2 balls from the box = 15C2 = (15 × 14) / 2

Required probability = (5 × 6) / ((15 × 14) / 2) = (5 × 6 × 2) / (15 × 14)   Ans. (b)


9. The results of the examinations in two papers A and B for 20 candidates were as follows: 

8 candidates passed in A , 7 candidates passed in B, 8  Candidates failed  in both. Out of these

Candidates one is selected. What is the probability he passed in both the papers?

(a)15/20          (b)3/20               (c) 4/20                 (d)17/20

Number of students failed in A = 20 - 8 = 12

Number of students failed in B = 13

Number of students failed in both = 8

Number of students failed in at least one paper = 12 + 13 - 8 = 17

                                                              (since failed in both is included in both A and B)

Number of students passed in both = 20 - 17 = 3

Required probability = Number of students passed in both / Total number of students = 3/20


10. From a pack of 52 playing cards 2 cards are drawn. What is the probability that it has one ace and

one jack?  (a)  4/663              (b)  8/663         (c) 16/663   (d) 32/663   

Since there are 4 Ace cards and 4 Jack cards, number of ways of drawing one ace and 

one Jack = 4 × 4 = 16

Number of ways of drawing 2 cards from the pack = 52C2 = (52 × 51) / 2 = 26 × 51

Required probability = 16 / (26 × 51) = 8 / (13 × 51) = 8 / 663


11. Two dice are thrown. What is the probability that the sum of numbers appeared is less than 8? 

(a) 3/12             (b) 5/12                (c) 7/12             (d)    11/12               

Ways of getting 7 from 2 dice are 1, 6; 2, 5; 3, 4; 4, 3; 5, 2; 6,1         6 ways

Ways of getting 6 from 2 dice are 1, 5; 2, 4; 3, 3; 4, 2; 5, 1                5 ways

Ways of getting 5 from 2 dice are 1, 4; 2, 3; 3, 2; 4,  1                       4 ways

Ways of getting 4 from 2 dice are 1, 3; 2, 2; 3, 1                                 3 ways

Ways of getting 3 from 2 dice are 1, 2; 2, 1                                         2 ways

Ways of getting 2 from 2 dice are 2, 2                                                  1 way

Ways of getting less than 8 from 2 dice = 1 + 2 + . . . + 6 = 21

Total possibilities of sum = 6 × 6

Required Prob. = 21 / (6 × 6) = 7 / 12


12. A group of 210 students appeared in a test. The mean of 1/3rd of students is found to be 60.

The mean of the remaining students is found to be 78. The mean of the whole group will be: 

(1) 80    (2) 76    (3)74    (4)72

Mean of the whole group = ⅓ × 60 + ⅔ × 78 = 20 + 52 = 72

Note that in the above calculation multiplying ⅓ and ⅔ by 210 and dividing the total sum by 210

and then cancel 210 in the numerator and the denominator. 

Hence it is not necessary to multiply ⅓ and ⅔ by 210 and divide the total sum by 210.


13. For the set of 500 observations variance is found to be 125. If 25 is added to each observation and 55 is subtracted then the new variance is.      (a)95            (b)150                       (c)180               (d)125

Adding or subtracting a constant to each value increases or decreases the mean by the same constant, but it will not affect the variance. (d)


14. Two students A and B scored the same average marks in 10 tests and the standard deviation of marks are 2.5 and 3 respectively for A and B. If the consistency of performance is the criterion for awarding prize, then

(a)B should get the prize       (b)A should get the prize    (c) Both A and B should share the prize equally 

    (d)Prize winner cannot be decided, since average mark is not given.

For consistency, deviations have to be minimum. A should get the prize


15. A cricketer has a certain average for 10 innings. In the eleventh innings, he scored 108  runs,

thereby increasing his average by 6 runs.

His new average is          (a)48     (b)42       (c)  52        (d)55 

108 runs in the 11th innings increased the total runs of 11 innings by 108. 

And it contribute to increase the average by 6 means that it increases the total of 10 innings by

6 × 10 = 60 to the new average. 

Hence 108 - 60 = 48  is the total of the remaining 11 - 10 = 1 innings to the new average.

Hence 48/1 = 48 is the new average.

Aliter :  If x is the new average then x - 6 is the average in the first 10 innings.

Total runs in the first 10 innings = 10(x - 6)

Hence 10(x - 6) + 108 = total runs in 11 innings = 11x

11x - 10x = 108 - 60

x = 48


16. There are 40 children in a class in which boys are 4 more than the girls. The average weight of all the students is 42.5 kg and the average weight of all the girls is 48 kg.

Then find the average weight of all the boys.

(a) 39.5 kg  (b) 38 kg   (c) 40.5 kg   (d) 36.75 kg   (e) 40.25 kg

Since the number of boys are 4 more than girls, Number of girls = (40 - 4) / 2 = 18

Number of boys = 40 - 18 = 22 

Total weight of all = 40 × 42.5 = 1700

Total weight of girls = 18 × 48 = 864

Total weight of boys = 1700 - 864 = 836

Average weight of all the boys = 836 / 22 = 418 / 11 = 38


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