Question

A life time of a certain component has a normal distribution with mean of 400 hours and standard deviation of 50 hours. Assuming a normal sample of 1000 components, find number of components whose life time lies between 340 to 465 hours.

[Given : A (z = 1.2) = 0.3849, A(z = 1.3) = 0.4032]

Answer 1

The number of components whose lifetime lies between 340 to 465 hours is 788.1.

Answer 2

The lifetime of the components follows a normal distribution with mean $\mu = 400$ hours and standard deviation $\sigma = 50$ hours. The sample size is $N = 1000$. We want to find the number of components whose lifetime lies between 340 and 465 hours.

First, we convert the given range of lifetimes (340 and 465 hours) into standard Z-scores using the formula $Z = \frac{X - \mu}{\sigma}$.

For $X_1 = 340$: $Z_1 = \frac{340 - 400}{50} = \frac{-60}{50} = -1.2$

For $X_2 = 465$: $Z_2 = \frac{465 - 400}{50} = \frac{65}{50} = 1.3$

We need to find the probability that a component's lifetime lies between 340 and 465 hours, which is $P(340 \le X \le 465)$. In terms of Z-scores, this is $P(-1.2 \le Z \le 1.3)$.

Using the properties of the standard normal distribution, this probability can be calculated as: $P(-1.2 \le Z \le 1.3) = P(-1.2 \le Z \le 0) + P(0 \le Z \le 1.3)$

Due to the symmetry of the normal distribution around the mean (Z=0), $P(-1.2 \le Z \le 0) = P(0 \le Z \le 1.2)$. We are given the area under the standard normal curve from 0 to z, denoted by $A(z) = P(0 \le Z \le z)$. Given $A(z = 1.2) = 0.3849$, so $P(0 \le Z \le 1.2) = 0.3849$. Given $A(z = 1.3) = 0.4032$, so $P(0 \le Z \le 1.3) = 0.4032$.

Therefore, the probability is: $P(-1.2 \le Z \le 1.3) = 0.3849 + 0.4032 = 0.7881$

This is the probability that a single component's lifetime lies between 340 and 465 hours.

To find the number of components in a sample of 1000 whose lifetime lies within this range, we multiply the total number of components by this probability: Expected number = Sample size $\times$ Probability Expected number = $1000 \times 0.7881 = 788.1$

Thus, the expected number of components whose lifetime lies between 340 and 465 hours is 788.1.

Explanation of the solution:

1. Identify mean ($\mu=400$) and standard deviation ($\sigma=50$).
2. Convert the range $[340, 465]$ to Z-scores: $Z_1 = \frac{340-400}{50} = -1.2$, $Z_2 = \frac{465-400}{50} = 1.3$.
3. Find the probability $P(-1.2 \le Z \le 1.3) = P(-1.2 \le Z \le 0) + P(0 \le Z \le 1.3)$.
4. Use symmetry $P(-1.2 \le Z \le 0) = P(0 \le Z \le 1.2) = A(1.2) = 0.3849$.
5. Use given $A(1.3) = 0.4032$.
6. Calculate probability: $0.3849 + 0.4032 = 0.7881$.
7. Calculate expected number: $1000 \times 0.7881 = 788.1$.