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

788.1

Answer 2

Let $X$ be the lifetime of a component, which follows a normal distribution with mean $\mu = 400$ hours and standard deviation $\sigma = 50$ hours.

We need to find the number of components whose lifetime lies between 340 and 465 hours from a sample of 1000 components.

First, we standardize the values 340 and 465 using the Z-score 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 the lifetime $X$ is between 340 and 465 hours, which is $P(340 \le X \le 465)$. This is equivalent to finding the probability that the Z-score is between -1.2 and 1.3, i.e., $P(-1.2 \le Z \le 1.3)$.

The area under the standard normal curve between $Z_1 = -1.2$ and $Z_2 = 1.3$ is given by: $P(-1.2 \le Z \le 1.3) = P(-1.2 \le Z \le 0) + P(0 \le Z \le 1.3)$

Due to the symmetry of the normal distribution about the mean (Z=0), $P(-1.2 \le Z \le 0) = P(0 \le Z \le 1.2)$. We are given the areas from the standard normal table: $A(z = 1.2) = P(0 \le Z \le 1.2) = 0.3849$ $A(z = 1.3) = P(0 \le Z \le 1.3) = 0.4032$

So, $P(-1.2 \le Z \le 1.3) = P(0 \le Z \le 1.2) + P(0 \le Z \le 1.3) = 0.3849 + 0.4032 = 0.7881$.

This is the probability that a single component's lifetime lies between 340 and 465 hours. The total number of components in the sample is 1000. The expected number of components whose lifetime lies between 340 and 465 hours is the total number of components multiplied by this probability: Expected number = $1000 \times P(340 \le X \le 465) = 1000 \times 0.7881 = 788.1$.

The number of components must be an integer in reality, but in probability calculations involving continuous distributions and large samples, the expected number can be a decimal. Following the approach of the similar question, we keep the decimal value.