Info given:With a random sample of 9 men. let X denote the number of men with head circumference larger than 57.8cm. the normal mean=56cm and standard dev=1.8cm.

So this is a binomal dis-

they ask;
For X, calculate pr(X=2)?

second part,
Let X denote the number of men with HC larger than 57.8cm in a random sample of size 100. Use normal Aprrox to calculate Pr(X<20)?

To calculate the probability Pr(X=2) for the first part, we need to use the binomial distribution formula. The binomial distribution describes the probability of getting a certain number of successes (in this case, men with head circumference larger than 57.8cm) given a fixed number of trials (the sample size) and a known probability of success.

In this case, X follows a binomial distribution with parameters n = 9 (sample size) and p = Pr(head circumference > 57.8cm). Since the mean and standard deviation are given, we can assume that the distribution is approximately normal.

To find p, we need to convert the head circumference cutoff value into a z-score using the given mean and standard deviation.

z = (x - μ) / σ
where x is the cutoff value, μ is the mean, and σ is the standard deviation.

In this case, z = (57.8 - 56) / 1.8 ≈ 0.7778.

Next, we need to find the probability of obtaining exactly 2 successes (men with head circumference larger than 57.8cm) out of 9 trials. We can use the binomial distribution formula:

Pr(X=k) = C(n, k) * p^k * (1-p)^(n-k)
where C(n, k) is the binomial coefficient (n choose k), p is the probability of success, and k is the number of successes.

Plugging in the values, we get:
Pr(X=2) = C(9, 2) * p^2 * (1-p)^(9-2)

The binomial coefficient C(9, 2) is calculated as:
C(9, 2) = 9! / (2!(9-2)!) = 9! / (2! * 7!) = 36.

Let's calculate p^2 * (1-p)^(9-2) next. Substituting the z-score value into the formula, we have:
p^2 * (1-p)^(9-2) ≈ (0.7778)^2 * (1-0.7778)^(9-2)

Evaluating this expression, we get:
(0.7778)^2 * (1-0.7778)^(9-2) ≈ 0.1481.

Finally, multiplying this with the binomial coefficient, we get:
Pr(X=2) ≈ 36 * 0.1481 ≈ 5.332.

Therefore, the probability Pr(X=2) is approximately 5.332.

For the second part, we are asked to use a normal approximation to calculate Pr(X<20), where X represents the number of men with head circumference larger than 57.8cm in a random sample of size 100.

Since the sample size is large (n = 100), according to the Central Limit Theorem, the binomial distribution can be approximated by a normal distribution with parameters μ = np and σ = sqrt(np(1-p)), where p is the probability of success.

Using the given p, we can calculate the mean and standard deviation of the normal distribution as:
μ = 100 * p
σ = sqrt(100 * p * (1-p))

To find Pr(X<20), we need to convert this into a z-score using the z-score formula:
z = (x - μ) / σ,
where x is the value of interest (in this case, 20).

Plugging in the values, we have:
z = (20 - μ) / σ.

Since we have the mean μ and standard deviation σ, we can calculate the z-score using this formula.

Finally, we can use a standard normal distribution table or software to find the probability Pr(Z < z), where Z is a standard normal random variable with mean 0 and standard deviation 1.

By looking up the corresponding z-score in the standard normal distribution table or using software, we can find Pr(Z < z). This would give us the desired probability Pr(X<20) using the normal approximation.

Note: In both parts, it is important to keep in mind that the normal approximation is an approximation and may not be as accurate as the exact binomial distribution, especially for small sample sizes.