Page 221 - DMTH404_STATISTICS

Unit 15: Exponential Distribution and Normal Distribution

Solution. Notes

Let X denote the diameter of the washer. Thus, X ~ N (5.02, 0.05).
The probability that a washer is defective = 1 - P(4.96 £ X £ 5.08)

-
-
éæ 4.96 5.02ö æ 5.08 5.02 ù ö
-
= 1 P ç ê ÷ £ z £ ç ÷ ú
ë è 0.05 ø è 0.05 ø û
-
z
-
= 1 P ( 1.2- £ z £ 1.2 ) = 1 2P ( 0 £ £ 1.2 ) = 1 - 2´ 0.3849 = 0.2302
Thus, the percentage of defective washers = 23.02.
Example 35: The average number of units produced by a manufacturing concern per day
is 355 with a standard deviation of 50. It makes a profit of Rs 1.50 per unit. Determine the
percentage of days when its total profit per day is (i) between Rs 457.50 and Rs 645.00, (ii) greater
than Rs 682.50 (assume the distribution to be normal). The area between z = 0 to z = 1 is 0.34134,
the area between z = 0 to z = 1.5 is 0.43319 and the area between z = 0 to z = 2 is 0.47725, where z
is a standard normal variate.
Solution.
Let X denote the profit per day. The mean of X is 355´ 1.50 = Rs 532.50 and its S.D. is
50´ 1.50 = Rs 75. Thus, X ~ N (532.50, 75).
(i) The probability of profit per day lying between Rs 457.50 and Rs 645.00

-
-
æ 457.50 532.50 645.00 532.50ö
P
P (457.50 £ X £ 645.00 ) = ç £ z £ ÷
è 75 75 ø
z
z
z
= P ( 1- £ £ 1.5 ) = P (0 £ £ ) 1 + P ( 0 £ £ 1.5 ) = 0.34134 + 0.43319 = 0.77453
Thus, the percentage of days = 77.453
-
æ 682.50 532.50ö
P z 
(ii) P (X  682.50 ) = ç ÷ = P (z  ) 2
è 75 ø
= 0.5000 P (0 £ £ ) 2 = 0.5000 0.47725 = 0.02275
-
z
-
Thus, the percentage of days = 2.275
Example 36:
The distribution of 1,000 examinees according to marks percentage is given below :

% Marks less than 40 40 - 75 75 or more Total
No. of examinees 430 420 150 1000

Assuming the marks percentage to follow a normal distribution, calculate the mean and standard
deviation of marks. If not more than 300 examinees are to fail, what should be the passing
marks?

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