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 Pr(X=2), where X denotes the number of men with head circumference larger than 57.8cm in a random sample of 9 men, we can use the binomial distribution formula.
The binomial distribution is defined by the formula:
Pr(X=k) = (n choose k) * p^k * (1-p)^(n-k)
where n is the sample size, k is the specific number of people with head circumference larger than 57.8cm, and p is the probability of success.
In this case, n = 9, k = 2, and p can be calculated using the normal mean and standard deviation provided. Since the head circumference is normally distributed, we can use z-scores to find the probability.
First, we need to calculate the z-score for a head circumference larger than 57.8cm using the formula:
z = (x - mean) / standard deviation
where x is the threshold value of 57.8cm, mean is the normal mean of 56cm, and the standard deviation is 1.8cm.
So, z = (57.8 - 56) / 1.8 = 0.889
Next, we need to find the cumulative probability up to this z-score. We can use a standard normal table or a calculator to find the corresponding probability.
Looking up the z-value of 0.889 in the standard normal table or using a calculator, we find that the cumulative probability (Pr(Z < 0.889)) is approximately 0.812.
Now, we can substitute the values into the binomial distribution formula:
Pr(X=2) = (9 choose 2) * (0.812^2) * (1 - 0.812)^(9-2)
Using the combination formula (9 choose 2) = 36, the calculated probability is:
Pr(X=2) = 36 * (0.812^2) * (1 - 0.812)^(9-2) ≈ 0.287 (rounded to three decimal places)
For the second part of the question, we need to find Pr(X<20) where X denotes the number of men with head circumference larger than 57.8cm in a random sample of size 100. In this case, we can use the normal approximation since the sample size is large enough.
We still need to calculate the z-score using the formula:
z = (x - mean) / standard deviation
In this case, x = 20, mean = 56, and standard deviation = 1.8. Calculating the z-score:
z = (20 - 56) / 1.8 ≈ -20
Since we want to find Pr(X<20), we need to calculate the cumulative probability up to this z-score. However, the standard normal table does not have a z-value of -20, as it is an extreme value.
To approximate this probability, we can use the property that the tail probabilities for extreme values approach zero. Therefore, Pr(X<20) will be a very small probability close to zero.
In practice, this means that Pr(X<20) is effectively zero, and we can conclude that Pr(X<20) ≈ 0.