Information given:the distribution of male head circumference (HC) is normal with a mean of 56cm and a standard deviation of 1.8cm

first question
-the prob that HC of a random chosen man is larger than 57.8cm is?
Work: pr(x)>57.8cm
z=(57.8-56)=1.8--->.9641 i got it wrong?

Second ques:
what is is the sampling distrubution of the avg HC of 9 random selected men?

Third: the prob that the average HC of 9 random selected men is larger then 57.8cm is?

First question: The probability that the head circumference of a randomly chosen man is larger than 57.8cm can be calculated using the standard normal distribution.

To begin, we can calculate the z-score using the formula z = (x - μ) / σ, where x is the value of interest (57.8cm), μ is the mean (56cm), and σ is the standard deviation (1.8cm).

So, the z-score is (57.8 - 56) / 1.8 = 0.8889.

Next, we can use the z-table or a calculator to find the probability associated with this z-score. Looking up the z-score in the table, we find that the probability is approximately 0.8128.

Therefore, the probability that the head circumference of a randomly chosen man is larger than 57.8cm is about 0.8128, or 81.28%.

Second question: The sampling distribution of the average head circumference of 9 randomly selected men can be approximated by a normal distribution with the same mean, but with a standard deviation equal to the population standard deviation divided by the square root of the sample size.

In this case, the mean of the sampling distribution remains 56cm, as it is the same as the population mean. The standard deviation of the sampling distribution can be calculated as σ / √n, where σ is the population standard deviation (1.8cm) and n is the sample size (9).

So, the standard deviation of the sampling distribution is 1.8 / √9 = 0.6cm.

Therefore, the sampling distribution of the average head circumference of 9 randomly selected men is a normal distribution with a mean of 56cm and a standard deviation of 0.6cm.

Third question: To find the probability that the average head circumference of 9 randomly selected men is larger than 57.8cm, we can again use the normal distribution.

We can calculate the z-score using the formula z = (x - μ) / (σ / √n), where x is the value of interest (57.8cm), μ is the mean (56cm), σ is the population standard deviation (1.8cm), and n is the sample size (9).

So, the z-score is (57.8 - 56) / (1.8 / √9) = 1.6667.

Using the z-table or a calculator, we find that the probability associated with this z-score is approximately 0.9522.

Therefore, the probability that the average head circumference of 9 randomly selected men is larger than 57.8cm is about 0.9522, or 95.22%.