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
Answer:
yes
Step-by-step explanation:
The temperature decreased so you would subtract the amount from the original temperature.
The answer would be:
-15 - 10 = -25
Low-birthweight i believe.
If the order doesn't matter, then the probability is 37.5%
If it has to be head head tail, then the probability is 12.5%
If the order doesn't matter, the possible outcomes that have two heads and one tail are HHT, HTH, and THH. Since these all have a probability of 12.5% of occuring, the probability of any of them occuring is 37.5%