Commuters were asked how many times a week they stop for coffee on their way to work. A survey found a mean of 3 times a week with a standard deviation of 0.55. What is the probability that in a group of 36 people they stop more than 3.2 times per week?

To find the probability that in a group of 36 people they stop more than 3.2 times per week, we need to use the concept of the sampling distribution of the mean.

The sampling distribution of the mean is the distribution of sample means obtained from repeated sampling from the same population. It follows a normal distribution when the sample size is sufficiently large (i.e., greater than or equal to 30) or when the population distribution is approximately normal.

In this case, we are given that the mean number of times commuters stop for coffee is 3 times per week, with a standard deviation of 0.55.

First, we need to calculate the standard deviation of the sampling distribution of the mean, also known as the standard error. The formula for the standard error is:

Standard Error = Standard Deviation / √(Sample Size)

Here, the standard deviation (σ) is 0.55 and the sample size (n) is 36.

Standard Error = 0.55 / √(36)
Standard Error = 0.55 / 6
Standard Error = 0.0916667 (rounded to 7 decimal places)

Now, we can calculate the z-score, which measures the number of standard deviations an observation or value is from the mean. The formula for the z-score is:

z = (X - μ) / SE

Here, X is the value we want to find the probability for, μ is the mean, and SE is the standard error.

Let's calculate the z-score for X = 3.2:

z = (3.2 - 3) / 0.0916667
z = 0.2 / 0.0916667
z = 2.181818 (rounded to 6 decimal places)

Using a standard normal distribution table or a calculator, we can find the probability associated with the z-score of 2.181818. The probability can be found by finding the area under the curve to the right of the z-score.

P(z > 2.181818) = 1 - P(z ≤ 2.181818)

By consulting a standard normal distribution table or using a calculator, we find that P(z ≤ 2.181818) ≈ 0.9857.

Therefore,

P(z > 2.181818) = 1 - 0.9857
P(z > 2.181818) ≈ 0.0143

So, the probability that in a group of 36 people they stop more than 3.2 times per week is approximately 0.0143, or 1.43%.