For women aged 18-24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1. a) ...

We are given a population (assumed to be approximately normally distributed) with a mean mu=114.8 and a standard deviation sigma=13.1

(a) We are asked to find the probability that a randomly chosen member of the population has a score greater than 120, or P(x>120).

We convert the raw score to a standard normal z-score:

z=(120-114.8)/13.1=0.40 (.3969465649)

We then consult a standard normal table (or use technology) to determine the area to the right of this...

We are given a population (assumed to be approximately normally distributed) with a mean mu=114.8 and a standard deviation sigma=13.1

(a) We are asked to find the probability that a randomly chosen member of the population has a score greater than 120, or P(x>120).

We convert the raw score to a standard normal z-score:

z=(120-114.8)/13.1=0.40 (.3969465649)

We then consult a standard normal table (or use technology) to determine the area to the right of this z-score under the standard normal curve. Many tables give the area to the left, so we would use the complement (subtract from 1.)

Using a TI-83/84 calculator we get P(x>120)=P(z>0.40)=.3457 (The table, assuming it gives the area to the left, will yield .6554; subtracting from 1 we get .3446)

Thus P(x>120) is approximately 35%

(b) If we want a group of 4 from this population to have a mean greater than 120, we employ much the same process. However, since we are talking about a sample mean, when computing the z-score we use the standard error which is sigma/sqrt(n), where n is the size of the sample.

P(bar(x)>120) ==> z=(120-114.8)/(13.1/sqrt(4))=.79 (.7938931298)

Now P(bar(x)>120)=P(z>0.79)=.2136 from a graphing utility. (A table will yield 1-.7852=.2148)

So P(bar(x)>120) is approximately 21%