Statistics, standard deviation help

[Calculator14]

Homework Statement

Suppose that the systolic blood pressure of 20 year old females is normally distributed with mean 120 and standard deviation 15.
a.) What proportion of 20 year old females will have readings between 115 and 125?
b.) If you selected a sample of 64 people from this population and then computed the mean; what is the probability that this mean value is between 115 and 125?

Homework Equations

(work shown below)

The Attempt at a Solution

a.)z= (115-120) / 15 = -0.333 P=.3707
z= (125-120) / 15 = 0.333 P=.6293
P(between 115 & 125) = .6293 - .3707 = 0.2586

b.) (15) / (64)^1/2 = 15/8 = 1.875
z= (115 – 120) / 1.875 = -2.6667 P=0.0038
z=(125 – 120) / 1.875 = 2.6667 P=0.9962
0.9962 – 0.0038 = 0.9924

 

[HallsofIvy]

Homework Statement

Suppose that the systolic blood pressure of 20 year old females is normally distributed with mean 120 and standard deviation 15.
a.) What proportion of 20 year old females will have readings between 115 and 125?
b.) If you selected a sample of 64 people from this population and then computed the mean; what is the probability that this mean value is between 115 and 125?

Homework Equations

(work shown below)

The Attempt at a Solution

a.)z= (115-120) / 15 = -0.333 P=.3707

This is actually for -0.33. If your table only gives two decimal places for z and you want better accuracy, you can interpolate between -0.33 and -0.34. -0.33 gives .3707 and -0.34 gives 0.3669. That's a difference of .3707- .3669= .0038 and .3 of that is 0.00114. Adding that back to .3669, we have 0.36804.

z= (125-120) / 15 = 0.333 P=.6293
P(between 115 & 125) = .6293 - .3707 = 0.2586

b.) (15) / (64)^1/2 = 15/8 = 1.875
z= (115 – 120) / 1.875 = -2.6667 P=0.0038
z=(125 – 120) / 1.875 = 2.6667 P=0.9962
0.9962 – 0.0038 = 0.9924