Arandstad/harris interactive survey reported that 25% of employees said their company isloyal to them (usa today, november 11, 2009). suppose 10 employees are selected randomlyand will be interviewed about company loyalty.
round your answers to four decimal places.
a. is the selection of ten employees a binomial experiment?
select: yes/no
explain.
the input in the box below will not be graded, but may be reviewed and considered by your instructor.
b. what is the probability that none of the 10 employees will say their company is loyal to them?
c. what is the probability that four of the 10 employees will say their company is loyal to them?
d. what is the probability that at least two of the 10 employees will say their company is loyal to them?

3-4

step-by-step explanation:

the answer is 3-4 because both 45 and 60 has a multiple of 15.

45/3=15

60/4=15

therefore the odds against selecting a red is 4-3

i can't solve unless i know what the line is

step-by-step explanation:

welcome to/p>

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;