Question

5. Thirty percent of all automobiles undergoing an emissions inspection at a certain inspection station fail the inspection. In a random sample of 15 selected cars: (round 3 decimal places) a) Find the probability that at most five of the cars fail the inspection. b) Find the probability that at least three of the cars fail the inspection. c) Find the probability that all 15 of the cars fail the inspection. 6. Twenty-five percent of the customers of a grocery store use an express checkout. In a random survey of ten customers: (round 3 decimal places) a) Find that probability that no more than two uses express checkout. b) Find the probability that at least 8 use express checkout. c) Find the probability that exactly half of the sample use express checkout. d) Find the expected number (mean) of the number that use express checkout. e) Find the standard deviation of the number that use express checkout.

Answer 1

5) 30% of all automobiles undergoing an emissions inspection fail inspection. This is same as the probability of a randomly selected car fails inspection is 0.30

Let X be the number of cars in a sample of 15 that fail the inspection. We can say that X has a Binomial distribution with parameters, number of trials (number of cars in the sample) n=15 and success probability ( the probability of a randomly selected car fails inspection) p=0.30

The probability of X=x cars fail the inspection is

$\begin{align*} P(X=x)&=\binom{n}{x}p^x(1-p)^{n-x}\\ &=\binom{15}{x}0.3^x(1-0.3)^{15-x},\quad x=0,1,\ldots,15\\ \end{align*}$

a) The probability that at most five (5 or less) cars fail the inspection is

$\begin{align*} P(X\le 5)&=P(X=0)+\ldots+P(X=5)\\ &=\binom{15}{0}0.3^0(1-0.3)^{15-0}+\ldots+\binom{15}{5}0.3^5(1-0.3)^{15-5}\\ &=\frac{15!}{0!(15-0)!}0.3^0(1-0.3)^{15-0}+\ldots+\frac{15!}{5!(15-5)!}0.3^5(1-0.3)^{15-5}\\ &=0.0047+ 0.0305+ 0.0916+ 0.17+ 0.2186+ 0.2061\\ &=0.7216 \end{align*}$

ans: The probability that at most five cars fail the inspection is 0.722

b) The probability that at least three (3 or more ) cars fail the inspection is

$\begin{align*} P(X\ge 3)&=1-P(X\le 2)\\ &=1-(P(X=0)+P(X=1)+P(X=2))\\ &=1-\left( \binom{15}{0}0.3^0(1-0.3)^{15-0}+\binom{15}{1}0.3^1(1-0.3)^{15-1}+\binom{15}{2}0.3^2(1-0.3)^{15-2}\right )\\ &=1-(0.0047+ 0.0305+ 0.0916)\\ &=0.8732 \end{align*}$

ans: The probability that at least three cars fail the inspection is 0.873

c) The probability that all 15 cars fail the inspection is

P(X = 15) = 15 0.315 (1 – 0.3) 5-15 15 15! 70.315(1 – 0.3)" 15!(15 – 15)! 0.0000

ans: The probability that all 15 cars fail the inspection is 0.000

6) 25% of the customers of a grocery store use an express checkout. This is same as, the probability of a randomly selected customer of the grocery store uses an express checkout is 0.25

Let X be the number of customers in a sample of 10 that use express checkout. We can say that X has a Binomial distribution with parameters, number of trials (number of customers) n=10 and success probability (the probability of a randomly selected customer of the grocery store uses an express checkout) p=0.25

The probability of X=x customers use express checkout is

2-T P(X = x) = (*)p*(1 – p)" - (1) 0.25" (1 – 0.25)10-1, 1 = 0,1, ..., 10

a) The probability that no more than 2 (2 or less) uses express checkout is

$\begin{align*} P(X\le 2)&=P(X=0)+P(X=1)+P(X=2)\\ &=\binom{10}{0}0.25^0(1-0.25)^{10-0}+\binom{10}{1}0.25^1(1-0.25)^{10-1}+\binom{10}{2}0.25^2(1-0.25)^{10-2}\\ &=0.0563+ 0.1877+ 0.2816\\ &=0.5256 \end{align*}$

ans: The probability that no more than 2 uses express checkout is 0.526

b) The probability that at least 8 (8 or more) uses express checkout is

$\begin{align*} P(X\ge 8)&=P(X=8)+P(X=9)+P(X=10)\\ &=\binom{10}{8}0.25^8(1-0.25)^{10-8}+\binom{10}{9}0.25^9(1-0.25)^{10-9}+\binom{10}{10}0.25^{10}(1-0.25)^{10-10}\\ &=0.0004+ 0+ 0\\ &=0.0004 \end{align*}$

ans: The probability that at least 8 uses express checkout is 0.000

c) The probability that exactly half of the sample (exactly 10/2=5) uses express checkout is

$\begin{align*} P(X=5)&=\binom{10}{5}0.25^5(1-0.25)^{10-5}\\ &=0.0584 \end{align*}$

ans: The probability that exactly half of the sample uses express checkout is 0.058

d) The expected value of X (using the formula for Binomial distribution)

$\begin{align*} E(X)=np=10\times 0.25=2.5 \end{align*}$

ans: The expected value of the number that use express checkout is 2.5

The standard deviation of X (using the formula for Binomial distribution)

$\begin{align*} \sigma=\sqrt{np(1-p)}=\sqrt{10\times 0.25\times (1-0.25)}=1.3693 \end{align*}$

ans: The standard deviation of the  number that use express checkout is 1.369