Question

Please Help! 50 points!!! (Dont spam my question ill report you)

Answer 1

Answer:

Step-by-step explanation:

1)The probability of success in each of the 58 identical engine tests is p=0.92

n = 58

mean, u = np = 58×0.92 = 53.36

2) The only value that would be considered usual for this distribution is 91. This is because it is the only value between the minimum and maximum value

3) n = 546

p = 17/100 = 0.17

Mean = np = 546×0.17= 92.82

4) n = 1035

p = 36/100 = 0.36

np = 1035 × 0.36 = 372.6

5) The probability of success is 0.2.

p = 0.2

q= 1-p = 1-0.2 = 0.8

n = 35

standard deviation =

√npq = √35×0.2×0.8 = 2.34

6) p = 0.25

q = 1-0.25 = 0.75

n = 5

Variance = npq = 5×0.25×0.75 = 0.9

7) n = 982

p = 0.431

q = 1 - p = 1 - 0.431 = 0.569

Variance = npq = 982×0.431×0.569= 240.8

8) n = 500

p = 84/100 = 0.84

q = 1-0.84 = 0.16

Standard deviation = √npq

Standard deviation = √500×0.84×0.16 = 8.2