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

Answer:

$P(\bar X>2.8)$

We can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z=\frac{2.8 -2.32}{\frac{1.24}{\sqrt{32}}}=2.190$

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

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

Step-by-step 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:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And replacing we got:

$z=\frac{2.8 -2.32}{\frac{1.24}{\sqrt{32}}}=2.190$

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

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

Answer 2

Answer:

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.

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.

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 2.32, \sigma = 1.24, n = 43, s = \frac{1.24}{\sqrt{43}} = 0.189$

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

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

By the Central Limit Theorem

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

$Z = \frac{2.8 - 2.32}{0.189}$

$Z = 2.54$

$Z = 2.54$ 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.