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<z<∞)-P(0<z<2.5)

From normal distribution table P(0<z<2.5)=0.4938 and we know that P(0<z<∞)=0.5. So,

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