Question

According to the American Consumer Satisfaction Index, hotels received an average satisfaction score of 67 in 2020 on a scale of 0-100. Assume that this satisfaction score is normally distributed with a standard deviation of8.2. What satisfaction score represents the 75th percentile?

Answer 1

To find the satisfaction score that represents the 75th percentile, we can use z-scores and the standard normal distribution table. The satisfaction score is approximately 72.51.

Step-by-step explanation:

To find the satisfaction score that represents the 75th percentile, we need to use the concept of z-scores and the corresponding standard normal distribution table. First, we calculate the z-score corresponding to the 75th percentile. Since the normal distribution is centered at a mean of 67 and has a standard deviation of 8.2, we can calculate the z-score using the formula:

z = (x - mean) / standard deviation

Substituting in the values, we get:

z = (x - 67) / 8.2

Next, we find the z-score associated with the 75th percentile from the standard normal distribution table, which is approximately 0.674. Solving the equation for x, we have:

0.674 = (x - 67) / 8.2

Multiplying both sides by 8.2 and rearranging the equation, we find:

x - 67 = 0.674 * 8.2

x - 67 = 5.5128

x = 67 + 5.5128

x ≈ 72.5128

Therefore, the satisfaction score that represents the 75th percentile is approximately 72.51.