This question is about statistics, specifically normal distribution. Here's how to solve it:

a) What is the probability that a customer will take more than 60 minutes?

First, we need to calculate the z-score. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is:

Z = (X - μ) / σ

where:
X is the value we are interested in (60 minutes in this case),
μ is the mean (45 minutes),
σ is the standard deviation (10 minutes).

So, Z = (60 - 45) / 10 = 1.5

We then look up this z-score in the z-table to find the probability. The value for 1.5 in the z-table is 0.9332. However, this is the probability that a customer will take 60 minutes or less. We want the probability that a customer will take more than 60 minutes, so we subtract the value from 1 (because the total probability is 1).

So, the probability that a customer will take more than 60 minutes is 1 - 0.9332 = 0.0668 or 6.68%.

b) What proportion of customers will take between 35 and 60 minutes?

We need to calculate the z-scores for both 35 and 60 minutes and find the probabilities for these z-scores.

For 35 minutes, Z = (35 - 45) / 10 = -1. The value for -1 in the z-table is 0.1587.

For 60 minutes, we already calculated the z-score and probability as 1.5 and 0.9332 respectively.

The proportion of customers who will take between 35 and 60 minutes is the difference between these two probabilities: 0.9332 - 0.1587 = 0.7745 or 77.45%.

c) From a sample of 36 customers, what is the probability that the sample mean will be less than 40 minutes?

When dealing with sample means, the standard deviation becomes σ/√n, where n is the sample size. In this case, n is 36, so the new standard deviation is 10/√36 = 10/6 = 1.67.

We then calculate the z-score for 40 minutes: Z = (40 - 45) / 1.67 = -3.

The value for -3 in the z-table is 0.0013 or 0.13%, so this is the probability that the sample mean will be less than 40 minutes.

Question

This question is about statistics, specifically normal distribution. Here's how to solve it:

a) What is the probability that a customer will take more than 60 minutes?

First, we need to calculate the z-score. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is:

Z = (X - μ) / σ

where:
X is the value we are interested in (60 minutes in this case),
μ is the mean (45 minutes),
σ is the standard deviation (10 minutes).

So, Z = (60 - 45) / 10 = 1.5

We then look up this z-score in the z-table to find the probability. The value for 1.5 in the z-table is 0.9332. However, this is the probability that a customer will take 60 minutes or less. We want the probability that a customer will take more than 60 minutes, so we subtract the value from 1 (because the total probability is 1).

So, the probability that a customer will take more than 60 minutes is 1 - 0.9332 = 0.0668 or 6.68%.

b) What proportion of customers will take between 35 and 60 minutes?

We need to calculate the z-scores for both 35 and 60 minutes and find the probabilities for these z-scores.

For 35 minutes, Z = (35 - 45) / 10 = -1. The value for -1 in the z-table is 0.1587.

For 60 minutes, we already calculated the z-score and probability as 1.5 and 0.9332 respectively.

The proportion of customers who will take between 35 and 60 minutes is the difference between these two probabilities: 0.9332 - 0.1587 = 0.7745 or 77.45%.

c) From a sample of 36 customers, what is the probability that the sample mean will be less than 40 minutes?

When dealing with sample means, the standard deviation becomes σ/√n, where n is the sample size. In this case, n is 36, so the new standard deviation is 10/√36 = 10/6 = 1.67.

We then calculate the z-score for 40 minutes: Z = (40 - 45) / 1.67 = -3.

The value for -3 in the z-table is 0.0013 or 0.13%, so this is the probability that the sample mean will be less than 40 minutes.

Knowee AI · Accepted Answer