Question

1. amazon prime is a subscription service for the online retailer amazon.com that gives the subscriber various perks such as free two - day shipping on most of the items sold by amazon. according to capitol one shopping, amazon prime members purchase a mean of 6.4 items per month, with a standard deviation of 2.3 items. assume the distribution of items purchased per month is approximately normal.

a. what is the probability that a randomly selected amazon prime member purchases at least ten items per month?
b. a researcher for amazon obtains the purchasing records for a random sample of 100 amazon prime members. he is interested in determining the number of amazon prime members that order at least ten items per month. define the random variable of interest t.
c. could the random variable defined in part b be considered a binomial random variable? explain.
d. find the probability that exactly five amazon prime members in the sample purchase at least ten items per month?

Explanation:

Step 1: Identify the distribution and parameters

We know that the number of items purchased per month, \( X \), follows a normal distribution with mean \( \mu = 6.4 \) and standard deviation \( \sigma = 2.3 \). We need to find \( P(X \geq 10) \). First, we calculate the z - score using the formula \( z=\frac{x-\mu}{\sigma} \).

For \( x = 10 \), the z - score is \( z=\frac{10 - 6.4}{2.3}=\frac{3.6}{2.3}\approx1.57 \)

Step 2: Find the probability using the z - table

We want \( P(X\geq10)=P(Z\geq1.57) \). Since the total area under the standard normal curve is 1, \( P(Z\geq1.57)=1 - P(Z < 1.57) \)

Looking up the value of \( P(Z < 1.57) \) in the standard normal table, we find that \( P(Z < 1.57)=0.9418 \)

Then \( P(Z\geq1.57)=1 - 0.9418 = 0.0582\) (approximate value, slight differences may occur due to table precision)

Answer:

The probability that a randomly selected Amazon Prime member purchases at least ten items per month is approximately \( 0.0582 \) (or \( 5.82\% \))