Arandomly selected customer support wait time (from a normal distribution) is calculated to be 1.8 standard deviations above its mean. what is the probability that another randomly selected customer wait time from the distribution will be less than 1.8 standard deviations from the mean?

a) 0.9641

b) 0.9976

c) 0.9281

d) 0.0359

72 fudge and 75 toffee

step-by-step explanation:

i hope this : )

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

i dont know

step-by-step explanation:

umm. i dont know