One study shows that 50% of all federal inmates are serving time for drug dealing. A random sample of 16 federal inmates is selected. What is the probablility that 3 or fewer are serving time for drug dealing?

10000000000000000000000000000000

i believe the answer is 68cm

step-by-step explanation:

i just took 34 x 2 because the space between v and y is the same as y to k.

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;