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according to the national retail federation, men spent an average of $484 over a recent black friday weekend. assume that the standard deviation for this population is $146. a random sample of 42 men shoppers over this weekend was selected. what is the probability that the average amount spent on black friday weekend from this sample was less than $500?

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The probability that the average amount spent on the Black Friday weekend from this sample of 42 men shoppers was less than $500 is approximately 0.7123 or 71.23%.

How to calculate the probability

To calculate the probability that the average amount spent on the Black Friday weekend from the sample of 42 men shoppers was less than $500, use the Central Limit Theorem and assume that the sampling distribution of the sample mean follows a normal distribution.

Given:

Population mean (μ) = $484

Population standard deviation (σ) = $146

Sample size (n) = 42

Since the sample size is large (n > 30), approximate the distribution of the sample mean as a normal distribution.

To calculate the probability, we need to standardize the sample mean using the formula for the standard error (SE):

SE = σ /
\sqrt(n)

SE = $146 /
\sqrt(42)

Next, calculate the z-score using the formula:

z = (x - μ) / SE

z = ($500 - $484) / ($146 / sqrt(42))

Now, look up the z-score in the standard normal distribution table or use statistical software to find the corresponding cumulative probability.

The probability that the average amount spent on the Black Friday weekend from this sample was less than $500 is the probability to the left of the calculated z-score.

Let's calculate the z-score:

z = (500 - 484) / (146 /
\sqrt(42))

z = 16 / (146 / 6.4807)

z ≈ 0.5604

Using a standard normal distribution table or statistical software, we find that the cumulative probability corresponding to a z-score of 0.5604 is approximately 0.7123.

Therefore, the probability that the average amount spent on the Black Friday weekend from this sample of 42 men shoppers was less than $500 is approximately 0.7123 or 71.23%.

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