Q:

If 65 percent of the population of a large community is in favor of a proposed rise in school taxes, approximate the probability that a random sample of 100 people will contain: (a) at least 50 who are in favor of the proposition;(b) between 60 and 70 inclusive who are in favor; (c) fewer than 75 in favor.

Accepted Solution

A:
Answer:  a) 0.99917  b)  0.7054  c) 0.9820Step-by-step explanation:Let X denotes the number of people are in favor of a proposed rise in school taxes.Era re given that p(probability that a person is in favor of a proposed rise )= 0.65Sample size : n= 100Mean= np = 100(0.65)=65Standard deviation: [tex]\sigma=\sqrt{np(1-p)}\\\\=\sqrt{100(0.65)(0.35)}\\\\=4.76969600708\approx4.77[/tex] a) Now, the probability that a random sample of 100 people will contain  at least 50 who are in favor of the proposition :[tex]P(x\geq50)=P(\dfrac{x-\mu}{\sigma}\geq\dfrac{50-65}{4.77})\\\\=P(z\geq-3.144)\\\\=P(z<3.144)=0.99917\ \ [\because P(Z>-z)=P(Z<z)][/tex]b) between 60 and 70 inclusive who are in favor;[tex]P(60<x<70)=P(\dfrac{60-65}{4.77}<\dfrac{x-\mu}{\sigma}<\dfrac{70-65}{4.77})\\\\=P(-1.048<z<1.048)\\\\=1-2P(z>1.048)\ \ [\because P(-z<Z<z)=1-2P(Z>|z|)]\\\\=1-2(0.1473193)=0.7053614\approx0.7054[/tex]c) fewer than 75 in favor.[tex]P(x<50)=P(\dfrac{x-\mu}{\sigma}<\dfrac{75-65}{4.77})\\\\=P(z<2.096)\\\\=0.9819589\approx0.9820\ [/tex]