Question

The mean percent of childhood asthma prevalence in 43 cities is 2.16%. A random sample of 31 of these cities is selected. What is the
probability that the mean childhood asthma prevalence for the sample is greater than 2.6% ? Interpret this probability. Assume that o=1.36%.

The probability is:

Answer 1

Probability that the mean childhood asthma prevalence for the sample is greater than 2.6% is 0.0357 or 3.57 %.

What is Probability ?

Probability is the measure of likelihood of an event.

For a Normal Distribution ,

the z-score formula is given by

$\rm Z = \dfrac{X - \mu }{\sigma}$

Here $\rm \mu$ is the mean and $\rm \sigma$ is the standard deviation .

It is the measure of the deviation of the data from its central tendency.

Each z-score has a probability value.

It is given in the question that

Mean of 2.16% =  $\rm \mu$ = 2.16

Standard deviation of 1.36% means that  $\rm \sigma$ = 1.36

For Sample of 31 the value of standard deviation is $\rm s = \dfrac{\sigma}{\sqrt{n}}$

s = 1.36/√31

s = 0.2442

Substituting the values

Z = (2.6 -2.16)/0.2442

Z =1.8

the p value from the graph of z and p , 0.9643

To determine value of probability  more than X is 1 - 0.9643= 0.0357

the

Probability that the mean childhood asthma prevalence for the sample is greater than 2.6% is 0.0357.

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