Question

Given a normal distribution with mean equals 102 and standard deviation equals 15 and given you select a sample of n=9 what is the probability that x bar is less than 95

Answer 1

If a variable of interest, X, has a normal distribution with mean μ and variance σ² and you take a random sample of the said variable and calculate its sample mean (Xbar), the sample mean will also be a random variable that shares the same type of distribution as the original variable. Then the sample mean Xbar has a normal distribution with mean μ and variance σ²/n

You can symbolize the distribution of the sample mean as follows: X[bar]~N(μ;σ²/n)

To determine the asked probability you have to work using the distribution of the sample mean.

P(Xbar<95)

You have to use the standard normal distribution to find the probability, to translate the value of Xbar to Z, you have to subtract the mean of the distribution and divide it by the standard deviation, using the following formula:

`Z=\frac{\text{Xbar-}\mu}{\frac{\sigma}{\sqrt[]{n}}}`

So, the first step is to calculate the Z value that corresponds to Xbar=95

Given that

μ=102

σ=15

n=9

`\begin{gathered} Z=\frac{95-102}{\frac{15}{\sqrt[]{9}}} \\ Z=(-7)/((15)/(3)) \\ Z=(-7)/(5) \\ Z=-1.4 \end{gathered}`

Both expressions are equal so that:

P(Xbar<95) = P(Z<-1.4)

Using the Z-table you can determine the probability of Z<-1.4

P(Z<-1.4)= 0.081

So the probability of Xbar being less than 95 is equal to 0.081 → 8.1%