Assume a binomial distribution. Then the mean of the distribution is $np$ , where $n$ is the number of people taken in a sample and $p$ is the probability of the mutation occurring in any given person, and the standard deviation is $\sqrt{np(1-p)}$ . Then in this case, the standard deviation is $\sqrt{600(0.06)(0.94)}\approx5.82$ .

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3 years ago

Among 46- to 51-year-olds, 28% say they have called a talk show while under the influence of peer pressurepeer pressure. Supp

sasho [114]

Answer:

0.8997, 0.9999

Step-by-step explanation:

Given that among 46- to 51-year-olds, 28% say they have called a talk show while under the influence of peer pressure.

i.e. X no of people who say they have called a talk show while under the influence of peer pressure is binomial with p = 0.28

Each person is independent of the other and there are only two outcomes

n =7

a) the probability that at least one has not called a talk showcalled a talk show while under the influence of peer pressure

= $P(Y\geq 1)$ where Y is binomial with p = 0.72

= $1-(1-0.72)^7\\=0.9999$

b) the probability that at least one has called a talk showcalled a talk show while under the influence of peer pressurepeer pressurethe probability that at least one has not called a talk showcalled a talk show while under the influence of peer pressure

= $P(x\geq 1) =1-P(0)\\= 1-(1-0.28)^7\\=0.8997$