A popular retail store knows that the purchase amounts by its customers is a random variable that follows a normal distribution with a mean of $30 and a standard deviation of $9.

What is the probability that a randomly selected customer will spend between $20 and $35 at this store?

Use same process as indicated in your following post.

To find the probability that a randomly selected customer will spend between $20 and $35 at this store, we need to standardize the values using Z-scores and then use the standard normal distribution table or calculator.

First, let's calculate the Z-score for both $20 and $35.

Z-score formula: Z = (X - μ) / σ

Where:
X = the value we want to calculate the Z-score for
μ = mean of the distribution
σ = standard deviation of the distribution

For $20:
Z = (20 - 30) / 9
Z = -1.11

For $35:
Z = (35 - 30) / 9
Z = 0.56

Next, we look up the Z-scores in the standard normal distribution table. The table shows the probabilities corresponding to the area under the standard normal curve.

The probability of a Z-score less than -1.11 is the same as the area to the left of -1.11. From the table, we find that the corresponding probability is 0.1335.

The probability of a Z-score less than 0.56 is the same as the area to the left of 0.56. From the table, we find that the corresponding probability is 0.7123.

Now, to find the probability of a customer spending between $20 and $35, we subtract the probability of a Z-score less than $20 (Z = -1.11) from the probability of a Z-score less than $35 (Z = 0.56).

P(20 ≤ X ≤ 35) = P(X ≤ 35) - P(X ≤ 20)
= 0.7123 - 0.1335
= 0.5788

Therefore, the probability that a randomly selected customer will spend between $20 and $35 at this store is approximately 0.5788 or 57.88%.