Answer:
Suppose the heights of 18-year-old men are approximately normally distributed, with mean 71 inches and standard deviation 4 inches.
(a) What is the probability that an 18-year-old man selected at random is between 70 and 72 inches tall? (Round your answer to four decimal places.)
z1 = (70-71)/4 = -0.25
z2 = (72-71/4 = 0.25
P(70<X<72) = p(-0.25<z<0.25) = 0.1974
Answer: 0.1974
(b) If a random sample of thirteen 18-year-old men is selected, what is the probability that the mean height x is between 70 and 72 inches? (Round your answer to four decimal places.)
z1 = (70-71)/(4/sqrt(13)) = -0.9014
z2 = (72-71/(4/sqrt(13)) = 0.9014
P(70<X<72) = p(-0.9014<z<0.9014) = 0.6326
Answer: 0.6326
please mark me the brainiest
Hey, me again! So, I'll be giving you the totals for EACH group.
For the first group, multiply 1/8 by 3 to get the answer;
3/8 lbs
Do the same for this group (and all of the others) so that the answers are:
Group 2 - 3/4lbs
Group 3 - 2/2 = 1lb
Group 4 - 5/8lb
Group 5 - 7/8
Answer:
so the answer is I don't know