Question

1.) Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of 30 months before certain electronic components deteriorate, causing the watch to become unreliable. The standard deviation of watch lifetimes is 10 months, and the distribution of lifetimes is normal.

(a) If Accrotime guarantees a full refund on any defective watch for 2 years after purchase, what percentage of total production will the company expect to replace?

(b) If Accrotime does not want to make refunds on more than 12% of the watches it makes, how long should the guarantee period be (to the nearest month)?

Step 1

(a) If Accrotime guarantees a full refund on any defective watch for 2 years after purchase, what percentage of total production will the company expect to replace?

We are asked to find the percentage of total production a company can expect to replace watches with an average life of 30 months and a standard deviation of watch lifetime of 10 months with a normal lifetime distribution.

First, to find probabilities and areas for a random variable x that follows a normal distribution with mean μ = 30 and standard deviation σ = 10, we must convert the given measurements x to z values. Recall that the formula

z =

x − μ

σ

is used to convert x values to z values.

The company will guarantee a full refund on any defective watch for 2 years after purchase. Since x is measured in months we need to first convert 2 years to 24 months. Then, we need to convert

x = 24 to a z value.

z

=

x − μ σ

=24-30/10

= -0.60

Step 2

Since the guarantee is up to 24 months (2 years) and the z value associated with x = 24 is z = −0.60, we need to find

P(z ≤ −0.60).

We will use a left-tail style table to determine the area, which gives cumulative areas to the left of a specified z. Since we are looking for the area to the left of a z value, we can read that value from the table directly.

Use the Standard Normal Distribution Table to find the area of the left of z = −0.60. This area gives us the probability P(z ≤ −0.60).

P(z ≤ −0.60) = 0.2743

Therefore, rounded to two decimal places, the company should expect to replace about 27.43% of its watches

Step 3

(b) If Accrotime does not want to make refunds on more than 12% of the watches it makes, how long should the guarantee period be (to the nearest month)?

We want to find the z value with 12% of the area under the standard normal curve to the left of z, since the company does not wish to refund more than 12%.

Since we are given the area in a left tail, we can use the Standard Normal Distribution Table directly to find z.Searching the interior of the table for the value 0.12, we see that the exact value is not in the table. Therefore, we must use approximation methods.

The value 0.12 will fall between the values 0.1210, which is associated with the z value ________, and the value 0.1190, which is associated with the z value ____________.

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3.) Suppose the heights of 18-year-old men are approximately normally distributed, with mean 70 inches and standard deviation 4 inches.

(a) What is the probability that an 18-year-old man selected at random is between 69 and 71 inches tall? (Round your answer to four decimal places.) ___________

(b) If a random sample of twelve 18-year-old men is selected, what is the probability that the mean height x is between 69 and 71 inches? (Round your answer to four decimal places.) __________

4.) In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

More than a decade ago, high levels of lead in the blood put 82% of children at risk. A concerted effort was made to remove lead from the environment. Now, suppose only 11% of children in the United States are at risk of high blood-lead levels.

(a) In a random sample of 208 children taken more than a decade ago, what is the probability that 50 or more had high blood-lead levels? (Round your answer to three decimal places.)
___________

(b) In a random sample of 208 children taken now, what is the probability that 50 or more have high blood-lead levels? (Round your answer to three decimal places.) _________

Answer 1