1. The owner of a fish market determined that the average weight for a catfish is \(3.4\) pounds with a standard deviation of \(0.8\) of a pound. A citation catfish should be one of the top \(2\%\) in weight. Assuming the weights of catfish are normally distributed, at what weight (in pounds) should the citation designation be established?

Let \(X =\) catfish weight. \(X\) has a normal distribution with mean \(3.4\) and standard deviation \(0.8\). Find the \(98^{th}\) percentile of \(X\).

qnorm(0.98,3.4,0.8)
[1] 5.042999
  1. A catalog company administered a survey to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to \(3.5\) minutes. What proportion of customers having to hold more than \(5.0\) minutes will hang up before placing an order?

Let \(X =\) how long customers are willing to wait. \(X\) has an exponential distribution with \(\lambda = \frac{1}{3.5}\). Find \(P(X>5)=1-P(X<5)\)

1-pexp(5,1/3.5)
[1] 0.239651
  1. The amount of time you have to wait at a dentist’s office before you are called in is uniformly distributed between zero and twenty minutes. What is the probability that you have to wait between 10 and 15 minutes?

Let \(X =\) time you have to wait at dentist’s office. \(X\) has a uniform distribution from \(0\) to \(20\), i.e. \(f(x) = \frac{1}{20}\) for \(0 \le x \le 20\)

  • \(P(10<X<15) = \frac{15-10}{20}=0.25\)
punif(15,0,20) - punif(10,0,20)
[1] 0.25
  1. Sales at a local plumbing wholesaler consist of both over-the-counter sales as well as deliveries. During the course of a month, over-the-counter sales have a mean of \(\$102,972\) with a standard deviation of \(\$13,523\). Over the same time period, deliveries average \(\$242,354\) with a standard deviation of \(\$24,956\). Assuming that the sales over-the-counter are independent of deliveries, what are the mean and standard deviation of the wholesaler monthly sales?

Let \(X =\) over the counter sales and let \(Y\) = deliveries and so \(X+Y\) = monthly sales.

  • \(\mu_{x+y} = \mu_x + \mu_y = 102972 + 242354 = 345326\)
102972 + 242354
[1] 345326
  • \(\sigma_{x+y} = \sqrt{\sigma_x^2 + \sigma_y^2} = \sqrt{13523^2 + 24956^2}=28384.39\)
sqrt(13523^2 + 24956^2)
## [1] 28384.39
  1. A random variable \(X\) has a mean of \(50\) and variance of \(50\), and a random variable \(Y\) has a mean of \(100\) and variance of \(200\). Given the random variables \(X\) and \(Y\) have a correlation coefficient equal to \(0.50\), find the mean and variance of the random variable \(W = 4X - 3Y\).

  • \(E(4X-3Y) = 4E(X) - 3E(Y) = 4 \cdot 50 - 3 \cdot 100 = -100\)
4*50 - 3*100
[1] -100
  • \(\sigma_{x-y} = \sqrt{4^2 \sigma_x^2 + (-3)^2 \sigma_y^2 + 2(4)(-3) \rho \sigma_x \sigma_y}=\sqrt{4^2 \cdot 50 + (-3)^2 \cdot 200 + 2(4)(-3)(0.5) \sqrt{50} \sqrt{200}}=37.42\)
sqrt(4^2 * 50 + (-3)^2 * 200 + 2*4*-3*.5*sqrt(50)*sqrt(200))
[1] 37.41657
  1. The profit for a production process is equal to \(\$7,200\) minus \(\$2.80\) times the number of units produced. The mean and variance for the number of units produced are \(1,100\) and \(800\), respectively. Find the mean and variance of the profit.

Let \(X =\) number of units produced with mean \(1100\) and standard deviation \(800\). Let \(Y\) = profit = \(7200 - 2.8X\)

  • \(\mu_y = 7200 - 2.8 \mu_x = 7200 - 2.8(1100)=4120\)
7200 - 2.8*1100
[1] 4120
  • \(\sigma_y^2 = 2.8^2 \sigma_x^2 = 2.8^2(800) = 6272\)
2.8^2*(800)
[1] 6272
  1. A cereal manufacturer produces a cereal that claims to contain 16 ounces in each box. A sample of boxes results in the following table.
Weight in Ounces 14 15 16 17
Probability 0.10 0.30 0.40 0.20

  • \(\mu_x = \sum xp(x) = 14(0.10) + 15(0.30) + 16(0.40) + 17(0.20) = 1.4+4.5+6.4+3.4 = 15.7\)
x <- c(14,15,16,17)
p <- c(0.1,0.3,0.4,0.2)
mean <- sum(x*p)
mean
[1] 15.7
  • \(\sigma_x = \sum (x - \mu_x)^2 p(x) = (14-15.7)^2(0.1) + (15-15.7)^2(0.3) + (16-15.7)^2(0.2) + (17-15.7)^2(0.2) = 0.9\)
x <- c(14,15,16,17)
p <- c(0.1,0.3,0.4,0.2)
mean <- sum(x*p)
dev.sq <- (x-mean)^2
sd <- sqrt(sum(dev.sq*p))
sd
[1] 0.9
  1. On average, an RV sales lot sells six RVs per month.

Let \(X =\) the number of RVs sold per month. \(X\) has a Poisson distribution with \(\lambda=6\).

  • \(P(X=4)=0.13\)
dpois(4,6)
[1] 0.1338526
  • \(P(X > 3)=1-P(X \le 3) = 0.85\)
1-ppois(3,6)
[1] 0.8487961
  1. In a shipment of 18 trucks to a local truck dealer, there are four trucks that don’t have air conditioning. Assume that you select four trucks at random.

Let \(X =\) trucks without air conditioning. \(X\) has a hypergeometric distribution with \(N=18\), \(S=4\), \(n=4\), \(x=0,1,2,3,4\)

  • \(P(X=2) = \frac{C_2^4 C_2^{14}}{C_4^{18}}=0.178\)
dhyper(2,4,14,4)
[1] 0.1784314
  • \(P(X \le 2) = P(X=0) + P(X=1) + P(X=2) = \frac{C_0^4 C_4^{14}}{C_4^{18}} + \frac{C_1^4 C_3^{14}}{C_4^{18}} + \frac{C_2^4 C_2^{14}}{C_4^{18}} = 0.327 + 0.476 + 0.178 = 0.981\)
x <- c(0,1,2)
p <- dhyper(x,4,14,4)
sum(p)
[1] 0.9813725
phyper(2,4,14,4)
[1] 0.9813725
  1. In a recent survey of high school students, it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of \(\$52.30\) and a standard deviation of \(\$18.23\).

Let \(\bar X =\) average amount that \(25\) randomly selected students spend on entertainment each week \(\sim N(52.30,\frac{18.23}{\sqrt{25}})\)

  • \(P(\bar X > 60) = 1 - P(\bar X < 60) = 0.017\)
1-pnorm(60,52.30,18.23/sqrt(25))
[1] 0.01734737
  • The \(35^{th}\) percentile of \(\bar X = 50.9\)
qnorm(0.35,52.3,18.23/sqrt(25))
## [1] 50.89512
  1. According to the Internal Revenue Service, \(80\%\) of all tax returns lead to a refund. A random sample of \(100\) tax returns is taken. What is the mean and standard error of the sampling distribution of the sample proportion of returns leading to refunds?

Let \(\hat p\) = sample proportion of the \(100\) tax returns that lead to refund.

\(\hat p\) has a normal distribution with mean \(p=0.8\) and standard error \(\sqrt{\frac{0.8(1-0.8)}{100}}=0.04\)

  • \(P(0.78 \le \hat p \le 0.84) = P(\hat p \le 0.84) - P(\hat p \le 0.78) = 0.5328\)
pnorm(0.84,0.8,0.04) - pnorm(0.78,0.8,0.04)
[1] 0.5328072
  • The \(40^{th}\) percentile of the distribution of \(\hat p\) is \(0.7899\)
qnorm(0.4,0.8,0.04)
## [1] 0.7898661
  1. The following table displays the joint probability distribution of two discrete random variables \(X\) and \(Y\)
\(x=1\) \(x=2\) \(x=3\)
\(y=0\) \(0.10\) \(0.12\) \(0.06\)
\(y=1\) \(0.05\) \(0.10\) \(0.11\)
\(y=2\) \(0.02\) \(0.16\) \(0.28\)

  • \(E(X) = \sum x p(x)= 1(0.17) + 2(0.38) + 3(0.45)=2.28\)
x <- c(1,2,3)
px <- c(.10+.05+.02,.12+.10+.16,.06+.11+.28)
meanx <- sum(x*px)
meanx
## [1] 2.28
  • \(E(Y|X=2) = \sum_y y p(y|x=2)= 0(\frac{12}{38}) + 1(\frac{10}{38}) + 2(\frac{16}{38}) = 1.11\)
y <- c(0,1,2)
pyx2 <- c(0.12,0.10,0.16)/(0.12+0.10+0.16)
meanyx2 <- sum(y*pyx2)
meanyx2
## [1] 1.105263
  • \(\sigma_{xy} = E(XY) - E(X)E(Y) = 0.25\)
y <- c(0,1,2)
py <- c(.1+.12+.06,.05+.1+.11,.02+.16+.28)
meany <- sum(y*py)
meany
## [1] 1.18
exy <- .05+2*.1+3*.11+2*.02+4*.16+6*.28
exy
## [1] 2.94
cov <- (exy-meanx*meany)
cov
## [1] 0.2496
  1. A basketball player makes \(80\) percent of his free throws during the regular season. Consider his next eight free throws.

Let \(X =\) number of free throws he makes in eight shots \(\sim Bin(8,0.80)\)

  • \(P(X \ge 6) = P(X=6)+P(X=7)+P(X=8) = 0.29 + 0.34 + 0.17 = 0.80\)
x <- c(6,7,8)
p <- dbinom(x,8,0.8)
x
[1] 6 7 8
p
[1] 0.2936013 0.3355443 0.1677722
sum(p)
[1] 0.7969178
  • \(\mu_x = np = 8(0.8) = 6.4\)
mean <- 8*0.8
mean
## [1] 6.4
  • \(\sigma_x = \sqrt{8 \cdot 0.8 \cdot (1-0.8)} = 1.13\)
sigma <- sqrt(8*0.8*0.2)
sigma
[1] 1.131371
  1. A cooler contains 14 drinks: 8 soft drinks and 6 beers. You select 3 drinks from the cooler. Let the random variable X be the number of soft drinks you get out of 3. Find the probability distribution function of X.

Let \(X =\) number of soft drinks you get. \(X\) has a hypergeometric distribution with \(N = 14\), \(S = 8\), \(n = 3\) and \(x = 0,1,2,3\).

\(x\) \(0\) \(1\) \(2\) \(3\)
\(p(x)\) \(\frac{C_0^8 C_3^6}{C_3^{14}}=0.05\) \(\frac{C_1^8 C_2^6}{C_3^{14}}=0.33\) \(\frac{C_2^8 C_1^6}{C_3^{14}}=0.46\) \(\frac{C_3^8 C_0^6}{C_3^{14}}=0.15\) 
x <- c(0,1,2,3)
p <- dhyper(x,8,6,3)
x
[1] 0 1 2 3
p
[1] 0.05494505 0.32967033 0.46153846 0.15384615