Answer to Question #307832 in Statistics and Probability for dasefe

Question

"An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed, with mean equal to 792 hours and a standard deviation of 62 hours. Find the probability that a random samples of 18 bulbs will have an average life between 794 and 807 hours."

Answer 1

We need to find P(794<X<807)

We will use the formula:

Z = $\frac{X-\mu}{\sigma/\sqrt{n}}$

Given $\mu$=792, $\sigma$= 62 and n = 18,

When X = 794

Z = $\frac{794-792}{62/\sqrt{18}}$= 0.14

Also,

When X = 807

Z= $\frac{807-792}{62/\sqrt{18}}$ = 1.03

Then;

P(794<X<807) = P(0.14<Z<1.03)

=P(Z<1.03) - P(Z<0.14)

From the normal distribution table,

P(Z<1.03) = 0.8485

and

P(Z<0.14) = 0.5557

Therefore,

P(794<X<807) = 0.8485 - 0.5557

= 0.2928

Answer: 0.2928