Answer to Question #283456 in Statistics and Probability for Md Altaf Raja

Question

The guaranteed average life of a certain type of electric light bulbs is 1000 hours with a standard deviation of 125 hours.

It is proposed to sample the output so as to assure that 90% of the bulbs do not fall short of the guaranteed average by

more than 2.5%. What should be the minimum size of the sample? (The area under standard normal curve from to

is 0.4)

Answer 1

Solution:

Let n be the size of the sample. Since the guaranteed mean is 1,000 we do not want the mean of the sample to be less than 2.5% of 1000 (i.e. 25 ) from 1000 so that it should not be below 1000-25=975. Hence $\bar{X}>975$ . It follows that

$|z|=\left|\dfrac{\bar{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\right|>\left|\dfrac{975-1000}{\dfrac{125}{\sqrt{n}}}\right|=\dfrac{\sqrt{n}}{5} .$

From the given condition, the area of the probability normal curve to the left of $\sqrt{n} / 5$ should be .9 or the area between 0 and $\sqrt{n} / 5$ is .4. As we are not worried about the bulbs which have life above the guaranteed life, it is a one-tailed problem.

From the table cf areas, the area between t=0 and t=1.281 is 4 .

$\therefore \quad \sqrt{n} / 5=1.281$ or n=41 approx.

The sample should not consist of less than 41 bulbs.