Assume that blood pressure readings are normally distributed with a mean of 120 and a standard deviation of 8. If 100 people are randomly selected, find the probability that their mean blood pressure will be greater than 122.

0.0062

Step-by-step explanation:

Let X be the random variable blood pressure. The information we are given are

mean blood pressure=μx=120

standard deviation blood pressure=σx=8

we have to find P (mean bp>122) = P (xbar>122).

P (Xbar>122) = P (xbar-μxbar/σxbar>122-μxbar/σxbar) = P (Z>122-μxbar/σxbar)

Now, we have to find μxbar and σxbar. We know that

μxbar=μx

σxbar=σx/sqrt (n).

So, μxbar=μx=120

σxbar=σx/sqrt (n) = 8/sqrt (100) = 0.8

P (Xbar>122) = P (Z>122-120/0.8) = P (Z>2.5) = P (0
From normal distribution table P (0
P (Xbar>122) = 0.5-0.4938=0.0062

There is 0.62% chance that the mean blood pressure of 100 people will be greater than 122