Question

Let {X1},{X2}, ...... ,{X16} be independent and identically distributed (iid) normally distributed random variables with mean 10 and standard deviation 2.
Determine P(8.9 ≤ X ≤ 11.46). (Use your z-tables)

Answer 1

Answer:

$P(\frac{8.9-10}{0.5} \leq Z \leq \frac{11.46-10}{0.5}) = P(-2.2 \leq Z \leq 0.73)$

And we can find this probability with this difference:

$P(\frac{8.9-10}{0.5} \leq Z \leq \frac{11.46-10}{0.5}) = P(-2.2 \leq Z \leq 2.92)=P(z<2.92) -P(Z<-2.2) = 0.998-0.0139 = 0.984$Step-by-step 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

For this case we assume that we have 16 observations each one independent and with the following distribution:

$X_i \sim N (\mu =10, \sigma =2)$

And for this case we assume that the question is:

$P(8.9 \leq \bar X \leq 11.46)$

And for this case since each random variable have a normal distribution we can conclude that the distriution for the sample man is given by:

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

And for this case we can find the standard error like this:

$\sigma_{\bar X}= \frac{2}{\sqrt{16}}=0.5$

And we can use the z score formula given by:

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

And using this formula we got:

$P(\frac{8.9-10}{0.5} \leq Z \leq \frac{11.46-10}{0.5}) = P(-2.2 \leq Z \leq 0.73)$

And we can find this probability with this difference:

$P(\frac{8.9-10}{0.5} \leq Z \leq \frac{11.46-10}{0.5}) = P(-2.2 \leq Z \leq 2.92)=P(z<2.92) -P(Z<-2.2) = 0.998-0.0139 = 0.984$