Question

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 798 hours and a standard deviation of 81 hours. Find the probability that a random samples of 47 bulbs will have an average life of greater than 826 hours

Answer 1

Mean of 47 normal distributed random variables with mean = 798 and standard deviation = 81 is normal distributed random variable X with mean $m = 798$ and standard deviation $\sigma = \frac{81}{\sqrt{47}} = 11.82$

Probability that X is greater than $X_0 = 826$ :

$Pr(X > X_0) = 1 - Pr(X < X_0) = 1 - \Phi(\frac{X_0 - m}{\sigma})$ , where $\Phi(x)$ is distribution function of normal standard random variable

$Pr(X > 826) = 1 - \Phi(\frac{826 - 798}{11.82}) = 0.009$

So, the probability that a random samples of 47 bulbs will have an average life of greater than 826 hours is approximately 0.9%.