Question

In the United States, the mean age of men when they marry for the first time follows the normal distribution with a mean of 24.6 years. The standard deviation of the distribution is 2.5 years. For a random sample of 63 men, what is the likelihood that the age when they were first married is less than 24.8 years? (Round your z value to 2 decimal places. Round your answer to 4 decimal places.)

Answer 1

Answer: 73,57%.

Explanation:

We need to find P(Z > 24.8)

To answer this question, we use the central limit theorem.

$Z = \frac{X - U}{\frac{S}{\sqrt{n} } }$

Where:

• X = Sample mean
• U = Population mean
• S = Population standard deviation
• n = sample size

Hence, replacing

Z = $\frac{24.8 - 24.6}{\frac{2.5}{\sqrt{63} } }$

Z = 0.63

We look up 0.63 on the normal distribution table, and we obtain 0.7357

Therefore, the probability that in a radom sample of 63 men, the mean marriage age is less than 24.8 years is 73,57%.