4.7 Review Questions
 Let X be the number of repair calls an appliance repair shop may receive during an hour. The probability distribution of X is given in the following table:
x 0 1 2 P(X=x) 2a 2a a  Find the value of a.
 Find the mean of X.
 Find the standard deviation of X.
 Are the events “receiving no more than one call” and “receiving two calls” mutually exclusive?
 Are the events “receiving no more than one call” and “receiving two calls” independent? Explain using calculations.
 Roll a balanced die four times,
 Find the probability of observing six at least once.
 Find the probability of observing six exactly once.
 Find the probability of observing six two to four times inclusively.
 How many times do we expect to observe a six?
 An insurance company wants to design a homeowner’s policy for midpriced homes. From data compiled by the company, it is known that the annual claim amount, X, in thousands of dollars, per homeowner is a random variable with the following probability distribution.
x 0 10 50 100 200 P(X=x) 0.95 0.045 0.004 0.0009 a  Determine the value of a.
 Find the expected annual claim amount per homeowner.
 Determine the expected annual claim amount for every 1000 homeowners.
 How much should the insurance company charge for the annual premium to average a net profit of $50 per policy?
 A sales representative for a tire manufacturer claims that the company’s steelbelted radials last at least 35,000 miles. A tire dealer decides to check that claim by testing eight of the tires. If 75% or more of the eight tires he tests last at least 35,000 miles, he will purchase tires from the sales representative. If, in fact, 90% of the steelbelted radials produced by the manufacturer last at least 35,000 miles, what is the probability that the tire dealer will purchase tires from the sales representative?
 From past experience, the owner of a restaurant knows that, on average, 4% of the parties that make reservations never show. How many reservations can the owner accept and still be at least 80% sure that all parties that make a reservation will show?
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 [latex]2a+2a+a=1\Longrightarrow 5a=1\Longrightarrow a=0.2.[/latex]
 [latex]P(X=0)=2a=0.4, P(X=1)=2a=0.4, P(x=2)=0.2.[/latex] [latex]\mu=\sum x P(X=x)=0\times 0.4+1\times 0.4+2\times 0.2=0.8.[/latex]
 [latex]\sigma=\sqrt{\sum x^2 P(X=x)\mu^2}=\sqrt{0^2\times 0.4+1^2\times 0.4+2^2\times 0.20.8^2}=0.74833.[/latex]
 Let A be the event of receiving no more than one call and B be the event of receiving two calls. The two events are mutually exclusive since they don’t overlap. That is P(A&B) = 0.
 Let A be the event of receiving no more than one call and B be the event of receiving two calls, then
[latex]P(A)=P(X\le 1)=P(X=0)+P(X=1)=0.4+0.4=0.8, \quad P(B)=P(X=2)=0.4.[/latex]The two events are NOT independent, since [latex]P(A\&B)=0\ne P(A)\times P(B)[/latex].

 Let X be the number of six, then X follows a binomial distribution with n = 4 and [latex]p=P(\mbox{rolling a six})=\frac{1}{6}[/latex]. We want
[latex]P(X\ge 1)=1P(X=0)=1_4C_0\left(\frac{1}{6}\right)^0\left(1\frac{1}{6}\right)^{40}=10.4823=0.5177.[/latex]  We want
[latex]P(X=1)=_4C_1\left(\frac{1}{6}\right)^1\left(1\frac{1}{6}\right)^{41}=0.3858.[/latex]  We want
[latex]\begin{aligned} P(2\le X\le 4)&=P(X=2)+P(X=3)+P(X=4)\\ &=_4C_2\left(\frac{1}{6}\right)^2\left(1\frac{1}{6}\right)^{42}+_4C_3\left(\frac{1}{6}\right)^3\left(1\frac{1}{6}\right)^{43}+_4C_4\left(\frac{1}{6}\right)^4\left(1\frac{1}{6}\right)^{44}\\ &=0.1157+0.0154+0.0008=0.1319.\\ \text{OR}&=1P(X=0)P(X=1)\\ &=1_4C_0\left(\frac{1}{6}\right)^0\left(1\frac{1}{6}\right)^{40}_4C_1\left(\frac{1}{6}\right)^1\left(1\frac{1}{6}\right)^{41}\\ &=10.48230.3858=0.1319.\end{aligned}[/latex]  [latex]\mu=np=4\times \frac{1}{6}=0.6667.[/latex]
 Let X be the number of six, then X follows a binomial distribution with n = 4 and [latex]p=P(\mbox{rolling a six})=\frac{1}{6}[/latex]. We want


[latex]\sum P(X=x)=1\Longrightarrow 0.95+0.045+0.004+0.0009+a=1\Longrightarrow a=10.9999=0.0001.[/latex]
 [latex]\mu=\sum x P(X=x)=0\times 0.95+10\times 0.045+50\times 0.004+100\times 0.0009+200\times 0.0001=0.76 (\mbox{\$1000}).[/latex]
 [latex]1000\times \mu=1000\times 0.76=760 (\mbox{\$1000}).[/latex]
 [latex]\mu+50=760+50=\$810.[/latex]

 Let X be the number of steelbelted radials lasting at least 35000 miles, then X follows a binomial distribution with n = 8 and p = 0.9. Since 75% of eight is 0.75 × 8 = 6, we want
[latex]\begin{aligned} P(X\ge 6)&=P(X=6)+P(X=7)+P(X=8)\\ &={}_8C_6(0.9)^6(10.9)^2+_8C_7(0.9)^7(10.9)^7+_8C_8(0.9)^8(10.9)^0\\ &=0.1488+0.3826+0.4305=0.9616.\end{aligned}[/latex]  Since 4% of the parties never show, 96% will show. We want
[latex]0.96^n\ge 0.8\Longrightarrow n\le\frac{\ln 0.8}{\ln 0.96}=5.47\Longrightarrow n=5.[/latex]