Question

Suppose that daily calorie consumption for american men follows a normal distribution with a mean of 2760 calories and a standard deviation of 500 calories.Suppose a health science researcher selects a random sample of 25 American men and records their calorie intake for 24 hours (1 day). Find the probability that the mean of her sample will be between 2700 and 2800 calories

Answer 1

Answer:

38.11% probability that the mean of her sample will be between 2700 and 2800 calories

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 2760, \sigma = 500, n = 25, s = \frac{500}{\sqrt{25}} = 100$

Find the probability that the mean of her sample will be between 2700 and 2800 calories

This is the pvalue of Z when X = 2800 subtracted by the pvalue of Z when X = 2700.

X = 2800

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{2800 - 2760}{100}$

$Z = 0.4$

$Z = 0.4$ has a pvalue of 0.6554

X = 2700

$Z = \frac{X - \mu}{s}$

$Z = \frac{2700 - 2760}{100}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

0.6554 - 0.2743 = 0.3811

38.11% probability that the mean of her sample will be between 2700 and 2800 calories