Question

The mean percent of childhood asthma prevalence in 43 cities is 2.32​%. A random sample of 32 of these cities is selected. What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%? Interpret this probability. Assume that sigmaequals1.24​%. The probability is nothing.

Answer 1

0.55% probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%. This means that a sample having an asthma prevalence of greater than 2.8% is unusual event, that is, unlikely.

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 = (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.

If X is more than two standard deviations from the mean, it is considered an unusual outcome.

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 = (\sigma)/(√(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 = 2.32, \sigma = 1.24, n = 43, s = (1.24)/(√(43)) = 0.189`

What is the probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%?

This is 1 subtracted by the pvalue of Z when X = 2.8. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (2.8 - 2.32)/(0.189)`

`Z = 2.54`

`Z = 2.54` has a pvalue of 0.9945

1 - 0.9945 = 0.0055

0.55% probability that the mean childhood asthma prevalence for the sample is greater than 2.8​%. This means that a sample having an asthma prevalence of greater than 2.8% is unusual event, that is, unlikely.

Answer 2

`P(\bar X>2.8)`

We can use the z score formula given by:

`z=(\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z=(2.8 -2.32)/((1.24)/(√(32)))=2.190`

And using the normal standard distribution and the complement rule we got:

`P(z>2.190 )= 1-P(z<2.190) = 1-0.986=0.014`

Explanation:

For this case w eknow the following parameters:

`\mu = 2.32` represent the mean

`\sigma =1.24` represent the deviation

n= 32 represent the sample sze selected

We want to find the following probability:

`P(\bar X>2.8)`

We can use the z score formula given by:

`z=(\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z=(2.8 -2.32)/((1.24)/(√(32)))=2.190`

And using the normal standard distribution and the complement rule we got:

`P(z>2.190 )= 1-P(z<2.190) = 1-0.986=0.014`