Find probability that a single randomly selected value is greater than 146.1. P(X > 146.1) and Find probability that a sample of size is randomly selected with a mean greater than 146.1. P(M > 146.1)

Question

Mean=141.2 and St. Deviation=27.6. Sample size n=116.

Answer 1

Hi Pamela,

I’m assuming your standard deviation is a population standard deviation, which means we can use Z since your sample size is large. For an individual value, we can use the classic equation in introductory statistics:

z=(x-mu)/sigma where:

x=single value you are given

mu=population mean

sigma=population standard deviation

Here,

x=146.1

mu=141.2

sigma=27.6

z=(146.1-141.2)/27.6

z=0.18

Now, go to the z table, find 0.1 at left, 0.08 at top.

This gives: P (Z<0.18)=0.5714

To get P(Z>0.18)=P(x<146.1), take:

1-P(Z<0.18)=

1-0.5714=0.4286

That answers the first question.

For the second question, we have a slightly different formula since we’re working with a sample:

z=(xbar-mu)/SE

xbar=sample mean

mu=population mean

SE=standard error, which has its own formula

SE=sigma/sqrt(n)

sigma=standard deviation

n=sample size

Here,

sigma=27.6

n=116

SE=27.6/sqrt(116)

SE=2.563

Returning to original formula:

z=(xbar-mu)/SE

xbar=141.1

mu=146.1

SE=2.563

z=(141.1-146.1)/2.563

z= -1.95

Fom Z-table,

P(Z< -1.95)=P(xbar<141.1)=0.0256

Now to get P(xbar>141.1), take:

1-P(xbar<141.1)

1-0.0256=0.9744

I hope this helps.