Find the probability that a sample of size n=180 is randomly selected with a mean greater than 151.2. P(X̅ > 151.2) =

Question

A population of values has a normal distribution with mean=164 and standard deviation=78.3. You intend to draw a random sample of size n=180.

Find the probability that a sample of size n=180 is randomly selected with a mean greater than 151.2.

P(X̅ > 151.2) = 

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Hi Kaitlyn,

This is a similar problem to the ones you posted earlier, just slightly different terminology. X-bar refers to a sample mean, which indicates that we need our old friend standard error. Recall:

SE=s/sqrt(n) where:

s=standard deviation

n=sample size

For this problem:

s=78.3

n=180

SE=78.3/sqrt(180)

SE=5.84

Now, we use our other old friend, z=(x-mu)/sigma, but substitute standard error in for sigma and x-bar in for x.

Thus, we have:

z=(x-bar-mu)/SE where:

x-bar=sample mean

mu=population mean

SE=standard error

Now, for this problem:

x-bar=151.2

mu=164

SE=5.84

Thus:

z=(151.2-164)/5.84

z=-2.19

From z-table--go to -2.1 in left column, 0.09 on right row:

P(Z<-2.19)=0.0143

Since we have a normal distribution:

P(Z<-2.19)=P(X-bar<151.2)=0.0143

But the question asks for greater than 151.2, so:

P(X>151.2)=1-P(x<151.2)=

1-0.0143=0.9857

P(X>151.2)=0.9857

I see you posted another question like this. See if you can work that one on your own; post again if you have any questions. Good luck.