Question

The body temperatures of all mosquitoes in a county have a mean of 57∘F and a standard deviation of 10∘F. What is the probability that in a sample of 25 mosquitoes the mean body temperature is greater than 59∘F, assuming the underlying distribution is normal? Do not write probability in terms of percentage. Round your answer to two decimal places.

Answer 1

Answer:

0.16 probability that in a sample of 25 mosquitoes the mean body temperature is greater than 59∘F, assuming the underlying distribution is normal.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, 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.

The body temperatures of all mosquitoes in a county have a mean of 57∘F and a standard deviation of 10∘F.

This means that $\mu = 57, \sigma = 10$

Sample of 25:

This means that $n = 25, s = \frac{10}{\sqrt{25}} = 2$

of 25 mosquitoes the mean body temperature is greater than 59∘F, assuming the underlying distribution is normal?

This is 1 subtracted by the pvalue of Z when X = 59. So

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

By the Central Limit Theorem

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

$Z = \frac{59 - 57}{2}$

$Z = 1$

$Z = 1$ has a pvalue of 0.84

1 - 0.84 = 0.16

0.16 probability that in a sample of 25 mosquitoes the mean body temperature is greater than 59∘F, assuming the underlying distribution is normal.