J. D. Power and Associates says that 60 percent of car buyers now use the Internet for research and price comparisons. (a)
Find the probability that in a sample of 8 car buyers, all 8 will use the Internet. (Round your answer to 4 decimal places.)

Probability

(b)
Find the probability that in a sample of 8 car buyers, at least 5 will use the Internet. (Round your answer to 4 decimal places.)

Probability

(c)
Find the probability that in a sample of 8 car buyers, more than 4 will use the Internet. (Round your answer to 4 decimal places.)

Probability

(d)
Find the mean and standard deviation of the probability distribution. (Round your mean value to 2 decimal places and standard deviation to 4 decimal places.)

Mean
Standard deviation

sgth thdhtxr ythdsx hfgh ysrth th srtrh

step-by-step explanation:

trshd h rdhsjsdrnytjdt jyjyukfddythj

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.