Question

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a population mean of 100 and a population standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample exceeds 104

Answer 1

`P( \bar X >104) = P(Z > (104-100)/((15)/(√(50)))) = P(Z>1.886)`

And we can use the complement rule and the normal standard distribution or excel and we got:

`P(z>1.886) = 1-P(Z<1.886) = 1-0.970 = 0.03`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the IQ of a population, and for this case we know the distribution for X is given by:

`X \sim N(100,15)`

Where
`\mu=100` and
`\sigma=15`

We select a sample of n = 50 and we want to find the probability that:

`P(\bar X >104)`

Since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And we can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And using this formula we got:

`P( \bar X >104) = P(Z > (104-100)/((15)/(√(50)))) = P(Z>1.886)`

And we can use the complement rule and the normal standard distribution or excel and we got:

`P(z>1.886) = 1-P(Z<1.886) = 1-0.970 = 0.03`