nformation 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:

To find the probability that the head circumference (HC) of a randomly chosen man is larger than 57.8cm, you need to calculate the z-score and then look up the corresponding probability in the normal distribution table.

1. Calculate the z-score:
z = (x - μ) / σ
where x is the value you want to find the probability for, μ is the mean, and σ is the standard deviation.

In this case,
x = 57.8cm
μ = 56cm
σ = 1.8cm

So, the z-score is:
z = (57.8 - 56) / 1.8 = 1.222

2. Look up the probability in the normal distribution table:
The probability you need to find is P(z > 1.222). Since the normal distribution is symmetric, find the probability for z < 1.222 and subtract it from 1.

Consulting the table, you can find that P(z < 1.222) is approximately 0.8888.

So, P(z > 1.222) = 1 - 0.8888 = 0.1112
Therefore, the probability that the HC of a randomly chosen man is larger than 57.8cm is approximately 0.1112, or 11.12%.

Second Question:

The sampling distribution of the average head circumference (HC) of a randomly selected sample of 9 men can be approximated by a normal distribution. The mean of this sampling distribution will be the same as the population mean because the sample size is large enough.

However, the standard deviation of the sampling distribution, also known as the standard error, can be calculated by dividing the population standard deviation by the square root of the sample size.

In this case, the population standard deviation (σ) is given as 1.8cm and the sample size (n) is 9.

So, the standard deviation of the sampling distribution (SE) is:
SE = σ / sqrt(n) = 1.8 / sqrt(9) = 1.8 / 3 = 0.6cm

Therefore, the sampling distribution of the average HC of 9 randomly selected men has a mean equal to the population mean (56cm) and a standard deviation of 0.6cm.

Third Question:

To find the probability that the average head circumference (HC) of 9 randomly selected men is larger than 57.8cm, you need to calculate the z-score and then look up the corresponding probability in the normal distribution table.

1. Calculate the z-score:
z = (x - μ) / SE
where x is the value you want to find the probability for, μ is the mean, and SE is the standard error.

In this case,
x = 57.8cm
μ = 56cm (population mean)
SE = 0.6cm (standard deviation of the sampling distribution)

So, the z-score is:
z = (57.8 - 56) / 0.6 = 3

2. Look up the probability in the normal distribution table:
The probability you need to find is P(z > 3).

Consulting the table, you can find that P(z > 3) is approximately 0.0013.

Therefore, the probability that the average HC of 9 randomly selected men is larger than 57.8cm is approximately 0.0013, or 0.13%.