1) Normal distribution with a mean of 110 and a standard deviation of 14. What score does someone need to achieve to be in the top 40%? Round up to the nearest whole number.

2. Amount people are willing to pay follows Normal distribution 6.06% of people willing to pay less than $73.75, and 11.90% of people willing to pay more than $99.49." find Mean and standard deviation.

3. The confidence interval for the sample proportion of 60 tech companies that fail in the first five years is (0.319, 0.481). What confidence level was used?

Answer:

1) 114

2) mean - 88.37, standard deviation = 9.43

3) 80%

Step-by-step explanation

1) z for top 40% is z such that P(Z < z) = 0.60. From the Z table, that value is 0.25

x = mean +z*standard deviation

x = 110 + 0.25(14) = 113.50

2) z such that 6.06% falls below is -1.55

z such that 11.90% falls above is 1.18

Those values are found using any Z table

mean - 1.55*standard deviation = 73.75

mean + 1.18*standard deviation = 99.49

subtract the equations to get -2.73*standard deviation = -25.74

standard deviation = (-25.74)/(-2.73) = 9.43

mean - 1.55(9.43) = 73.75

mean = 73.75 + 1.55(9.43) = 88.37

3) p^ = (.481 + .319)/2 = .40

p^ + z*sqrt(p^(1-p^)/n) = .481

.40 + z*sqrt(.4*.6/60) = .481

0.63246z = .481 - .40

.063246z = .081

z = 1.28

80% interval