Suppose I know that female height is normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
What is the probability of picking a woman less than 70 inches tall?
0.9772
0.0228
0.1359
0.1587
Solution :
Given that ,
mean =
= 64
standard deviation =
= 3
P(X< 70) = P[(X-
) /
< (70-64) /3 ]
= P(z <2 )
Using z table
=0.9772
Suppose I know that female height is normally distributed with a mean of 64 inches and...
Suppose that height is normally distributed with a mean of 68 inches and a standard deviation of 4 inches. How many inches tall are the tallest 3% of people?
Assume that the height of men are normally distributed with a mean
of 69.8 inches and a standard deviation deviation of 3.5 inches. If
100 men are randomly selected, find thr probability that they have
a mean height greater than 69 inches.
Asume that the heights of men are normally distributed with a mean of 69.8 inches and a standard deviation of 3.5 inches of 100 men wa randomly selected in the probability that they have a meaning greater than...
The height of women ages 20-29 is normally distributed, with a mean of 63.963.9 inches. Assume sigmaσequals=2.62.6 inches. Are you more likely to randomly select 1 woman with a height less than 64.764.7 inches or are you more likely to select a sample of 1717 women with a mean height less than 64.764.7 inches? Explain. LOADING... Click the icon to view page 1 of the standard normal table. LOADING... Click the icon to view page 2 of the standard normal...
The height of women ages 20-29 is normally distributed with a mean of 64.1 inches. Assume o = 24 inches. Are you more likely to randomly select 1 woman with a height less than 65.5 inches or are you more likely to select a sample of 22 women with a mean height less than 65.5 inches? Explain Click the icon to view page 1 of the standard normal table Click the icon to view page 2 of the standard normal...
The height of women ages 20-29 is normally distributed, with a mean of 64.5 inches. Assume σ-2.9 inches. Are you more likely to randomly select 1 woman with a height less than 66.7 inches or are you more likely to select a sample of 15 women with a mean height less than 66.7 inches? Explain Click the icon to view page 1 of the standard normal table Click the icon to view page 2 of the standard normal table. What...
The height of women ages 20-29 is normally distributed, with a mean of 64.4 inches. Assume σ=2.4 inches. Are you more likely to randomly select 1 woman with a height less than 66.5 inches or are you more likely to select a sample of 15 women with a mean height less than 66.5 inches? Explain. (a) What is the probability of randomly selecting 1 woman with a height less than 66.5 inches?
The average height of American females aged 18-24 is normally distributed with mean u 64.4 inches and o 2.2 inches. Complete parts a. and b. below. a. What percentage of females are taller than 68 inches? What integral gives the probability that a female is taller than 68 inches? dX (Type exact answers, using as needed.) The percentage of females that are taller than 68 inches is (Round to the nearest whole number as needed.) b. What is the probability...
Assume the heights in a female population are normally distributed with mean 65.7 inches and standard deviation 3.2 inches. Then the probability that a typical female from this population is between 5 feet and 5 feet 7 inches tall is (to the nearest three decimals) which of the following? a. 0.620 b. 0.658 c. 0.963
Men heights are assumed to be normally distributed with mean 70 inches and standard deviation 4 inches; What is the probability that 4 randomly selected men are all less than 72 inches in height?
The heights of adult men in America are normally distributed, with a mean of 69.1 inches and a standard deviation of 2.69 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.6 inches and a standard deviation of 2.55 inches. a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? b) If a woman is 5 feet 11 inches tall, what is her z-score...