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

Answer:

$z =\frac{18.45-19}{\frac{7}{\sqrt{106}}}= -0.809$

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

$P(z<-0.809) = 0.209$

Step-by-step 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:

$\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})$

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:

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

And replacing we got:

$z =\frac{18.45-19}{\frac{7}{\sqrt{106}}}= -0.809$

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

$P(z<-0.809) = 0.209$