Question

Rochester, New York averages μ=21.9 inches of snow for the month of December. The distribution of snowfall amounts is approxiamtely normal with a standar deviation of σ=6.5 inches. This year a local jewelry store is advertising a refund of 50% off all purchases made in December, if the accumulated snowfall by the end of the month falls within the top 10% of the distribution. What is the probability that the jewelry store will have to pay off on its promise? How many inches of snow actually have to fall for the jewelry store to provide a refund?

Answer 1

Let X inches be the snow fall for the month of December in Rochester., New York. X has a normal distribution with mean and standard deviation .

The local jewelry store is advertising a refund of 50% off all purchases made in December, if the accumulated snowfall by the end of the month falls within the top 10% of the distribution. This is same as, the probability that the snow fall is with in top 10% of the distribution is 0.10.

ans: the probability that the jewelry store will have to pay off on its promise is 0.10

Let q inches be the amount of snow actually have to fall for the jewelry store to provide a refund.

This is same as the probability that amount of snow is greater than q inches is 0.10.

Let us first get the z value at which

Using the standard normal table, we can get for z=1.28, P(Z<1.28) = 0.90

That mean,

P(Z>1.28) = 0.10

However, we need

We can equate the z score of q to 1.28 and get

ans: 30.22 inches of snow actually have to fall for the jewelry store to provide a refund.