Suppose you know the distribution of sample proportions of female employees has a mean of 0.42 and a standard deviation of 0.21. Suppose you select a random sample of employees and find that the proportion of women in the sample is 0.45. | What is the standard score (z-score) for this sample?

| What is the probability that a second sample selected would have a proportion greater than 0.45?

-2x + 7 = 7

subtract 7 from both sides.

-2x = 0

divide both sides by -2.

x = 0

answer: 80,000

step-by-step explanation:

the population doubles every day.

10,000*2=20,000 day 4

20,000*2=40,000 day 5

40,000*2=80,000 day 6

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560