Question

Let X be the lifetime of an electronic device. It is known that the average lifetime of the device is 767 days and the standard deviation is 121 days. Let x¯ be the sample mean of the lifetimes of 157 devices. The distribution of X is unknown, however, the distribution of x¯ should be approximately normal according to the Central Limit Theorem. Calculate the following probabilities using the normal approximation. (a) P(x¯≤754)= Answer (b) P(x¯≥784)= Answer 0.0392 (c) P(749≤x¯≤784)= Answer

Answer 1

X be the lifetime of an electronic device

average lifetime of the device : = 767 days

Standard deviation : = 121

Sample size : n= 157

According to the Central Limit Theorem Sample mean follows normal distribution with mean: 767 days and standard deviation: =9.6569

(a) P(754)

Z-score for 754 = (754 - 767)/9.6569 = -1.35

From standard normal tables ; P(Z-1.35) = 0.0885

Ans : P(754) = 0.0885

(b)

P(784) = 1 - P(784)

Z-score for 784 = (784 - 767)/9.6569 = 1.76

From Standard normal tables ;

P(Z1.76) = 0.9608

P(784) = P(Z1.76) = 0.9608

P(784) = 1 - P(784) = 1-0.9608=0.0392

Ans : P(784) = 0.0392

(c)

P(749784) = P(784) - P(749)

From (b) P(784)=0.9608

Z-score for 749 = (749 - 767)/9.6569 = -1.86

From standard normal tables:

P(Z-1.86) = 0.0314

P(749) = P(Z-1.86) = 0.0314

P(749784) = P(784) - P(749) = 0.9608-0.0314=0.9294

Ans : P(749784) = 0.9294