Question

Suppose that we will randomly select a sample of 106 measurements from a population having a mean equal to 19 and a standard deviation equal to 7. Calculate the probability that we will obtain a sample mean less than 18.450; that is, calculate P( x < 18.450).

Answer 1

`z =(18.45-19)/((7)/(√(106)))= -0.809`

And if we use the normal standard distribution or excel we got:

`P(z<-0.809) = 0.209`

Explanation:

For this case we have the following info given:

`\mu = 19` represent the mean

`\sigma = 7` represent the standard deviation

`n = 106` represent the sample size

The distribution for the sample size if we use the central limit theorem (n>30) is given by:

`\bar X \sim N(\mu , (\sigma)/(√(n)))`

And for this case we want to find the following probability:

`P(\bar X< 18.45)`

And for this case we can use the z score formula given by:

`z =(\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z =(18.45-19)/((7)/(√(106)))= -0.809`

And if we use the normal standard distribution or excel we got:

`P(z<-0.809) = 0.209`