Bookstore has a website for which 90 percent of their book sales are to out of state customers and not subject to the collection of sales tax . The sample proportion of all sales to out of state customers from samples of size 100 will vary in repeated sampling. The sampling distribution for p is aproximately normal with a mean of 0.9 and s.deviation of 0.03. Find probabilities that between of 0.87 and 0.93 are out of state customers,that at least probability of .93 are out of state customers, that most 0.81 is out of state and that least 0.88 out of state ?

To find the probabilities for various scenarios related to the proportion of out of state customers buying books from the website, we can use the normal distribution and the given mean and standard deviation.

1. Probability that between 0.87 and 0.93 are out of state customers:
To find the probability that the proportion of out of state customers falls between 0.87 and 0.93, we need to find the area under the normal curve between these two values. This can be calculated using the cumulative distribution function (CDF) of the normal distribution.

P(0.87 ≤ p ≤ 0.93) = P(p ≤ 0.93) - P(p ≤ 0.87)
Using the Z-table or a calculator with the normal distribution function, we can find the corresponding probabilities for each value and subtract them:

P(p ≤ 0.93) ≈ P(Z ≤ (0.93 - 0.9) / 0.03)
P(p ≤ 0.93) ≈ P(Z ≤ 1)

P(p ≤ 0.87) ≈ P(Z ≤ (0.87 - 0.9) / 0.03)
P(p ≤ 0.87) ≈ P(Z ≤ -1)

By looking up the values in the standard normal distribution table, we find that P(Z ≤ 1) is approximately 0.8413 and P(Z ≤ -1) is approximately 0.1587.

Therefore, P(0.87 ≤ p ≤ 0.93) ≈ 0.8413 - 0.1587 = 0.6826

So, there is approximately a 68.26% probability that the proportion of out of state customers falls between 0.87 and 0.93.

2. Probability that at least 0.93 are out of state customers:
To find the probability that the proportion of out of state customers is at least 0.93, we need to calculate the area under the normal curve to the right of this value.

P(p ≥ 0.93) ≈ 1 - P(p ≤ 0.93)
P(p ≥ 0.93) ≈ 1 - P(Z ≤ (0.93 - 0.9) / 0.03)
P(p ≥ 0.93) ≈ 1 - P(Z ≤ 1)

Using the standard normal distribution table, we find that P(Z ≤ 1) is approximately 0.8413.

Therefore, P(p ≥ 0.93) ≈ 1 - 0.8413 = 0.1587

So, there is approximately a 15.87% probability that the proportion of out of state customers is at least 0.93.

3. Probability that at most 0.81 is out of state customers:
To find the probability that the proportion of out of state customers is at most 0.81, we need to calculate the area under the normal curve to the left of this value.

P(p ≤ 0.81) ≈ P(Z ≤ (0.81 - 0.9) / 0.03)
P(p ≤ 0.81) ≈ P(Z ≤ -3)

Using the standard normal distribution table, we find that P(Z ≤ -3) is approximately 0.0013.

Therefore, P(p ≤ 0.81) ≈ 0.0013

So, there is approximately a 0.13% probability that the proportion of out of state customers is at most 0.81.

4. Probability that at least 0.88 is out of state customers:
To find the probability that the proportion of out of state customers is at least 0.88, we need to calculate the area under the normal curve to the right of this value.

P(p ≥ 0.88) ≈ 1 - P(p ≤ 0.88)
P(p ≥ 0.88) ≈ 1 - P(Z ≤ (0.88 - 0.9) / 0.03)
P(p ≥ 0.88) ≈ 1 - P(Z ≤ -0.67)

Using the standard normal distribution table, we find that P(Z ≤ -0.67) is approximately 0.2514.

Therefore, P(p ≥ 0.88) ≈ 1 - 0.2514 = 0.7486

So, there is approximately a 74.86% probability that the proportion of out of state customers is at least 0.88.