The average age of residents in a large residential retirement community is 69 years with standard deviation 5.8 years. A simple random sample of 100 residents is to be selected, and the sample mean age Xof these residents is to be computed. The probability that the average age, Xof the 100 residents selected is less than 68.5 years is:

0.194

Step-by-step explanation:

X has a mean of 69 and a standard deviation of 5.8/√100 = 0.58.

The z-score is:

z = (x − μ) / σ

z = (68.5 − 69) / 0.58

z = -0.862

The probability is:

P(Z < -0.862) = 0.194

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