10. Suppose that a poll taken 10 years ago found that 65% of parents spank their
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Question
10. Suppose that a poll taken 10 years ago found that 65% of parents spank their children. Suppose a recent poll of 150 adults with children finds that 80 indicated that they spank their children. Complete parts (a) and (b) below.
(a) Assuming parents' attitude toward spanking has not changed since the original poll, how many of the 150 parents surveyed would be expected to spank their children?
The expected amount of parents surveyed that would spank their children is ___?__
(Round to the nearest whole number as needed.)
9. According to an almanac, 80% of adult smokers started smoking before turning 18 years old. When technology is used, use the Tech Help button for further assistance.
(a) Compute the mean and standard deviation of the random variable X, the number of smokers who started before 18 in 400 trials of the probability experiment.
(b) Interpret the mean.
(c) Would it be unusual to observe 360 smokers who started smoking before turning 18 years old in a random sample of 400 adult smokers? Why?
(a) x = __?__
Explanation / Answer
10 a) Expectd = n*p = 150*0.65 = 97.5 = 96 parents
9) a) Mean = n*p = 400*0.8 = 320
b) Expect 320 out of a sample of 400 adults who smoke to have started smoking before they turned 18
c) SD = sqrt(n*p*(1-p)) = sqrt(400*0.8*0.2) = 8
Formula: Z = (X - mean)/SD,
Using continuity correction also,
P(X > 360) = P(Z > (360.5 - 320)/8) = P(Z > 5.0625) = 0
It would be unusual as the probability is much less than 0.05(significance level)
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