Question

Assuming you obtained a random sample of blood glucose levels from 32 individuals from a population and the mean of the population is 120 and the standard deviation of the sample is 15, calculate the probability that the mean blood glucose levels for the sample population will be above 127.

Answer 1

Answer:

0.41% probability that the mean blood glucose levels for the sample population will be above 127.

Step-by-step explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using 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 problem, we have that:

$\mu = 120, \sigma = 15, n = 32, s = \frac{15}{\sqrt{32}} = 2.65$

Calculate the probability that the mean blood glucose levels for the sample population will be above 127.

This is 1 subtracted by the pvalue of Z when X = 127. So

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

By the Central Limit Theorem

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

$Z = \frac{127 - 120}{2.65}$

$Z = 2.64$

$Z = 2.64$ has a pvalue of 0.9959

1 - 0.9959 = 0.0041

0.41% probability that the mean blood glucose levels for the sample population will be above 127.