Question

the length of time taken on the sats follows a normally distribution with a mean of 2.5 hours and a standard deviation of 0.25 hours. samples of size of 60 are randomly selected from the population.
· x is the normal distribution of the length of time taken on a sat by a student. μx = 2.5 and σx = 0.25
· n = 60. since x is the normal distribution and n = 60, by clt, we can conclude that x follows a normal distribution.
· x ~ n(2.5, 0.25 / √60)
part b:
find the 95th percentile for a sample mean hours.
(round to two decimal places.)

Explanation:

Step1: Find the z - score for the 95th percentile

The z - score \(z\) corresponding to the 95th percentile (\(P(Z\leq z)=0.95\)) can be found using the standard normal distribution table or a calculator. Using a standard normal table or a calculator (e.g., in R: qnorm(0.95) or in Excel: NORM.S.INV(0.95)), the \(z\) - score is \(z = 1.645\).

Step2: Use the formula for the value of a sample mean in a normal distribution

The formula for the sample mean \(\bar{X}\) in a normal distribution is \(\bar{X}=\mu_{\bar{X}}+z\times\frac{\sigma_{X}}{\sqrt{n}}\).
We know that \(\mu_{\bar{X}} = 2.5\), \(z = 1.645\), \(\sigma_{X}=0.25\), and \(n = 60\).
First, calculate \(\frac{\sigma_{X}}{\sqrt{n}}=\frac{0.25}{\sqrt{60}}\approx\frac{0.25}{7.746}\approx0.0323\).
Then, \(\bar{X}=2.5 + 1.645\times0.0323\).
\(\bar{X}=2.5+ 0.0531\).

Answer:

\(2.55\)