Ch. 6.1 #7-49 odd

153.9 175.5
The z value is z = 175.5 159.3 1.88
8.64
From table E (Normal Distribution ), z=1.88 → 0.9699
1 – 0.9699 = 0.0301 or 3.01%
8. Household Computers According to recent surveys, 60% of households have personal computers. If a
random sample of 180 households is selected, what is the probability that more than 60 but fewer than
100 have a personal computer?
n=180, p=0.60
q = 1–q = 1 – 0.60 = 0.40
Since np = (180)(0.60) = 108 and nq = (180)(0.40) = 72, the normal distribution can be used to
approximate the binomial distribution (np ≥ 5 and nq ≥ 5).
µ = np = (180)(0.60) = 108
σ = npq = (180)(0.60)(0.40) 6.57
For binomial distribution, P(60 x100)
For normal distribution, P(60.5 < X < 99.5)
= P(60.5 < X < 99.5)
z = 99.5 108 1.29 area=0.0985
6.57
z = 60.5 108 7.23 area=0.0001
6.57
P(60.5 < X < 99.5)= P(-7.23 < z< -1.29)=0.0985-0.0001=0.0984 or 9.84%
0.4955 + 0.4945 = 0.99 or 0.99%