Question 1049353: I really need assistance with this question:
Men’s weights are normally distributed with mean 170 pounds and standard deviation 20 pounds.
(a) Find the probability that a randomly selected man has a weight between 140 and 190. (Show work and round the answer to 4 decimal places)
(b) What is the 80th percentile for men’s weight? (Show work and round the answer to 2 decimal places)
(c) If 64 men are randomly selected, find the probability that sample mean weight is greater than 175. (Show work and round the answer to 4 decimal places)
I really appreciate any help, thank you!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Men’s weights are normally distributed with mean 170 pounds and standard deviation 20 pounds.
(a) Find the probability that a randomly selected man has a weight between 140 and 190. (Show work and round the answer to 4 decimal places)
z(140) = (140-170)/20 = -3/2
z(190) = (190-170)/20 = 1
P(140< x < 190) = P(-3/2< z < 1) = normalcdf(-3/2,1) = 0.7745
-----------------------------
(b) What is the 80th percentile for men’s weight? (Show work and round the answer to 2 decimal places)
Find the z-value with a left-tail of 0.80:: invNorm(0.8) = 0.8416
That means the 80th percentile value is 0.8416 standard deviations
above the mean.
Find that value using x = z*s + u
Ans: x = 0.8416*20 + 170 = 186.83 lbs
-----------------------------------
(c) If 64 men are randomly selected, find the probability that sample mean weight is greater than 175. (Show work and round the answer to 4 decimal places)
mean of sample mean = 170
std of sample means = 20/sqrt(64) = 20/8 = 5/2
z(175) = (175-170)/(5/2) = 5/(5/2) = 2
P(x > 175) = P(z > 2) = normalcdf(2,100) = 0.0228
------------
Cheers,
Stan H.
---------------
|
|
|
