Question

Assume that the game playthrough times for a newly released puzzle game has a mean of 49.8 minutes and a standard deviation of 4.2 minutes. A sample of size n39 playthrough times is randomly selected from a gaming population. What is the probability that the sample.mean is in between 50 minutes and 51 minutes? You may use a calculator or the portion of the z-table given below. Round your answers to three decimal places where appropriate

Answer 1

The probability that the sample mean is in between 50 minutes and 51 minutes is 0.345.

How to calculate probability?

n = sample size = 39

σ = standard deviation =4.2

μ = population mean = 49.8

Probability is the chance that an event can occur, in this case probability is used to measure the chance sample mean is between 50 minutes and 51 minutes. So,

P(50 ≤ x ≤ 51) = P($\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$ ≤ z ≤ $\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$)

= P($\frac{50-49.8}{\frac{4.2}{\sqrt{39}}}$ ≤ z ≤ $\frac{51-49.8}{\frac{4.2}{\sqrt{39}}}$)

= P(0.30 ≤ z ≤ 1.78)

= P(z ≤ 1.78) - P(z ≤ 0.3)

using z table and we get,

= 0.9625 - 0.6179

= 0.3446

= 0.345

Thus, the probability is 0.345 for the sample mean is within 50 and 51.

Learn more about probability here:

brainly.com/question/29490792

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