Chapter 4: Distributions of random variablesContent check — Statistics (OpenIntro), Chapter 4: Distributions of random variables

$N(\mu =5,\sigma =3)$⁠.

$N(\mu =-100,\sigma =10)$⁠.

$N(\mu =2,\sigma =9)$⁠.

$Z_{Tom}=$ $\frac{x_{\text{Tom}}-{\mu }_{\text{ACT}}}{{\sigma }_{\text{ACT}}}=$ $\frac{24-21}{6}=$ $0.5$

Its Z-score is given by $Z=$ $\frac{x-\mu }{\sigma }=$ $\frac{5.19-3}{2}=$ $2.19/2=$ $1.095$⁠.

The observation $x$ is 1.095 standard deviations above the mean. We know it must be above the mean since $Z$ is positive.

For $x_1=95.4$ mm: $Z_1=$ $\frac{x_1-\mu }{\sigma }=$ $\frac{95.4-92.6}{3.6}=$ $0.78$⁠. For $x_2=85.8$ mm: $Z_2=\frac{85.8-92.6}{3.6}=-1.89$⁠.

Because the absolute value of Z-score for the second observation is larger than that of the first, the second observation has a more unusual head length.

We found this probability in Example 4.7: 0.6736.

If Edward did better than 36% of SAT takers, then about 64% must have done better than him.

We leave the drawings to you.

$Z=\frac{1500-1100}{200}=2\to 0.9772$⁠.

$1-0.9772=0.0228$⁠.

First put the heights into inches: 67 and 76 inches. Figures are shown below.

$Z_{\text{Mike}}=$ $\frac{67-70}{3.3}=$ $-$ $0.91\to$ $0.1814$⁠.

$Z_{\text{Jose}}=$ $\frac{76-70}{3.3}=$ $1.82\to$ $0.9656$⁠.

Short answers:

$Z_{95}=1.6449\to 1429$ SAT score.

$Z_{97.5}=1.96\to 1492$ SAT score.

Short answers:

$Z=1.21\to 0.8869$⁠, then subtract this value from 1 to get 0.1131.

$Z=-0.30\to 0.3821$⁠.

This is an abbreviated solution. (Be sure to draw a figure!) First find the percent who get below 1100 and the percent that get above 1400: $Z_{1100}=0.00\to 0.5000$ (area below), $Z_{1400}=1.5\to 0.0668$ (area above). Final answer: $1.0000-$ $0.5000-$ $0.0668=$ $0.4332$⁠.

5’5” is 65 inches ($Z=-1.52$⁠). 5’7” is 67 inches ($Z=-0.91$⁠). Numerical solution: $1.000-$ $0.0643-$ $0.8186=$ $0.1171$⁠, i.e. 11.71%.

First draw the pictures. Using software, we get 0.6827 within 1 standard deviation, 0.9545 within 2 standard deviations, and 0.9973 within 3 standard deviations.

700 and 1500 represent two standard deviations below and above the mean, which means about 95% of test takers will score between 700 and 1500.

We found that 700 to 1500 represents about 95% of test takers. These test takers would be evenly split by the center of the distribution, 1100, so $\frac{95\%}{2}=47.5\%$ of all test takers score between 1100 and 1500.

8.85%.

6.94%.

58.86%.

4.56%.

Verbal: $N(\mu =151,\sigma =7)$⁠, Quant: $N(\mu =153,\sigma =7.67)$⁠.

$Z_{VR}=1.29$⁠, $Z_{QR}=0.52$⁠.

She scored 1.29 standard deviations above the mean on the Verbal Reasoning section and 0.52 standard deviations above the mean on the Quantitative Reasoning section.

She did better on the Verbal Reasoning section since her Z-score on that section was higher.

$Per{c}_{VR}=0.9007\approx 90\%$⁠, $Per{c}_{QR}=0.6990\approx 70\%$⁠.

$100\%-90\%=10\%$ did better than her on VR, and $100\%-70\%=30\%$ did better than her on QR.

We cannot compare the raw scores since they are on different scales. Comparing her percentile scores is more appropriate when comparing her performance to others.

Answer to part (b) would not change as Z-scores can be calculated for distributions that are not normal. However, we could not answer parts (d)-(f) since we cannot use the normal probability table to calculate probabilities and percentiles without a normal model.

$Z=0.84$⁠, which corresponds to approximately 159 on QR.

$Z=-0.52$⁠, which corresponds to approximately 147 on VR.

$Z=1.2$⁠, $P(Z>1.2)=0.1151$⁠.

$Z=-1.28\to {70.6}^{\circ }$⁠F or colder.

$N(25,2.78)$⁠.

$Z=1.08$⁠, $P(Z>1.08)=0.1401$⁠.

The answers are very close because only the units were changed. (The only reason why they differ at all because 28${}^{\circ }$C is 82.4${}^{\circ }$F, not precisely 83${}^{\circ }$F.)

Since $IQR=Q3-Q1$⁠, we first need to find $Q3$ and $Q1$ and take the difference between the two. Remember that $Q3$ is the ${75}^{th}$ and $Q1$ is the ${25}^{th}$ percentile of a distribution. Q1 = 23.13, Q3 = 26.86, IQR = 26. 86 - 23.13 = 3.73.

We would expect to see about $1/0.7\approx 1.43$ individuals to find the first success.

First find the probability of the complement: $P($⁠no success in first 3 trials⁠$)={0.3}^3=0.027$⁠. Next, compute one minus this probability: $1-P($⁠no success in 3 trials⁠$)=1-0.027=0.973$⁠.

No. The geometric distribution is always right skewed and can never be well-approximated by the normal model.

No. The cards are not independent. For example, if the first card is an ace of clubs, that implies the second card cannot be an ace of clubs. Additionally, there are many possible categories, which would need to be simplified.

No. There are six events under consideration. The Bernoulli distribution allows for only two events or categories. Note that rolling a die could be a Bernoulli trial if we simplify to two events, e.g. rolling a 6 and not rolling a 6, though specifying such details would be necessary.

${0.875}^2\times 0.125=0.096$⁠.

$\mu =8$⁠, $\sigma =7.48$⁠.

If $p$ is the probability of a success, then the mean of a Bernoulli random variable $X$ is given by
$\mu =$ $E[X]=$ $P(X=0)\times 0+$ $P(X=1)\times 1$
$=(1-p)\times 0+p\times 1=0+p=p$

$P(A=$ $\text{not},$ $$ $B=$ $\text{exceed},$ $$ $C=$ $\text{not},$ $$ $D=$ $\text{not})=$ $(0.7)(0.3)(0.7)(0.7)=$ ${(0.7)}^3{(0.3)}^1$⁠.

We are asked to determine the expected number (the mean) and the standard deviation, both of which can be directly computed from the formulas: $\mu =np=40\times 0.7=28$ and $\sigma =$ $\sqrt{np(1-p)}=$ $\sqrt{40\times 0.7\times 0.3}=$ $2.9$⁠. Because very roughly 95% of observations fall within 2 standard deviations of the mean (see Section 2.1.4), we would probably observe at least 22 but fewer than 34 individuals in our sample who would not exceed the deductible.

One possible answer: if the friends know each other, then the independence assumption is probably not satisfied. For example, acquaintances may have similar smoking habits, or those friends might make a pact to quit together.

To check if the binomial model is appropriate, we must verify the conditions. (i) Since we are supposing we can treat the friends as a random sample, they are independent. (ii) We have a fixed number of trials ($n=4$⁠). (iii) Each outcome is a success or failure. (iv) The probability of a success is the same for each trials since the individuals are like a random sample ($p=0.3$ if we say a “success” is someone getting a lung condition, a morbid choice). Compute parts

and

using the binomial formula: $P(0)=$ $\left(\frac{4}{0}\right){(0.3)}^0{(0.7)}^4=$ $1\times 1\times {0.7}^4=$ $0.2401$⁠, $P(1)=$ $\left(\frac{4}{1}\right){(0.3)}^1{(0.7)}^3=$ $0.4116$⁠. Note: $0!=1$⁠. Part

can be computed as the sum of parts (a) and (b): $P(0)+$ $P(1)=$ $0.2401+$ $0.4116=$ $0.6517$⁠. That is, there is about a 65% chance that no more than one of your four smoking friends will develop a severe lung condition.

The complement (no more than one will develop a severe lung condition) as computed in Guided Practice 4.33 as 0.6517, so we compute one minus this value: 0.3483.

$\mu =0.3\times 7=2.1$⁠.

$P($⁠0, 1, or 2 develop severe lung condition⁠$)=$ $P(k=0)+$ $P(k=1)+$ $P(k=2)=$ $0.6471$⁠.

Frame these expressions into words. How many different ways are there to arrange 0 successes and $n$ failures in $n$ trials? (1 way.) How many different ways are there to arrange $n$ successes and 0 failures in $n$ trials? (1 way.)

One success and $n-1$ failures: there are exactly $n$ unique places we can put the success, so there are $n$ ways to arrange one success and $n-1$ failures. A similar argument is used for the second question. Mathematically, we show these results by verifying the following two equations:

$$\begin{array}{c}\hfill \left(\frac{n}{1}\right)=n,\left(\frac{n}{n-1}\right)=n\end{array}$$

The distribution is transformed from a blocky and skewed distribution into one that rather resembles the normal distribution in last hollow histogram.

Compute the Z-score first: $Z=\frac{42-60}{7.14}=-2.52$⁠. The corresponding left tail area is 0.0059.

Binomial conditions are met: (1) Independent trials: In a random sample, whether or not one 18-20 year old has consumed alcohol does not depend on whether or not another one has. (2) Fixed number of trials: $n=10$⁠. (3) Only two outcomes at each trial: Consumed or did not consume alcohol. (4) Probability of a success is the same for each trial: $p=0.697$⁠.

0.203.

0.203.

0.167.

0.997.

$\mu =35$⁠, $\sigma =3.24$

$Z=\frac{45-35}{3.24}=3.09$⁠. 45 is more than 3 standard deviations away from the mean, we can assume that it is an unusual observation. Therefore yes, we would be surprised.

Using the normal approximation, 0.0010. With 0.5 correction, 0.0017.

$1-{0.75}^3=0.5781$⁠.

0.1406.

0.4219.

$1-{0.25}^3=0.9844$⁠.

Geometric distribution: 0.109.

Binomial: 0.219.

Binomial: 0.137.

$1-{0.875}^6=0.551$⁠.

Geometric: 0.084.

Using a binomial distribution with $n=6$ and $p=0.75$⁠, we see that $\mu =4.5$⁠, $\sigma =1.06$⁠, and $Z=2.36$⁠. Since this is not within 2 SD, it may be considered unusual.

$\stackrel{Anna}{<mn>1</mn><mi>/</mi><mn>5</mn>}\times \stackrel{Ben}{<mn>1</mn><mi>/</mi><mn>4</mn>}\times \stackrel{Carl}{<mn>1</mn><mi>/</mi><mn>3</mn>}\times \stackrel{Damian}{<mn>1</mn><mi>/</mi><mn>2</mn>}\times \stackrel{Eddy}{<mn>1</mn><mi>/</mi><mn>1</mn>}=$ $1/5!=$ $1/120$⁠.

Since the probabilities must add to 1, there must be $5!=120$ possible orderings.

$8!=\text{40,320}$⁠.

One possible answer: since he is likely to make each field goal attempt, it will take him at least 4 attempts but probably not more than 6 or 7.

The first sequence: $0.2\times 0.2\times 0.8\times 0.8\times 0.8\times 0.8=$ $0.0164$⁠.

Answers may vary. We cannot conclusively say they are or are not independent. However, many statistical reviews of athletic performance suggests such attempts are very nearly independent.

If his fourth field goal ($k=4$⁠) is within five attempts, it either took him four or five tries ($n=4$ or $n=5$⁠). We have $p=0.8$ from earlier. Use the negative binomial distribution to compute the probability of $n=4$ tries and $n=5$ tries, then add those probabilities together:

$$\begin{array}{cc}\hfill & P(n=4\text{ OR }n=5)=P(n=4)+P(n=5)\hfill \\ \hfill & =\left(\frac{4-1}{4-1}\right){0.8}^4+\left(\frac{5-1}{4-1}\right){(0.8)}^4(1-0.8)=1\times 0.41+4\times 0.082=0.41+0.33=0.74\hfill \end{array}$$

In each part, $p=0.7$⁠.

The number of days is fixed, so this is binomial. The parameters are $k=3$ and $n=7$⁠: 0.097.

The last “success” (admitting a heart attack patient) is fixed to the last day, so we should apply the negative binomial distribution. The parameters are $k=2$⁠, $n=4$⁠: 0.132.

This problem is negative binomial with $k=1$ and $n=5$⁠: 0.006. Note that the negative binomial case when $k=1$ is the same as using the geometric distribution.

Geometric, 0.0804.

Binomial, 0.0322.

Negative binomial, 0.0193.

Negative binomial with $n=4$ and $p=0.55$⁠, where a success is defined here as a female student. The negative binomial setting is appropriate since the last trial is fixed but the order of the first 3 trials is unknown.

0.1838.

$\left(\frac{3}{1}\right)=3$⁠.

In the binomial model there are no restrictions on the outcome of the last trial. In the negative binomial model the last trial is fixed. Therefore we are interested in the number of ways of orderings of the other $k-1$ successes in the first $n-1$ trials.

Poisson with $\lambda =75$⁠.

$\mu =\lambda =75$⁠, $\sigma =\sqrt{\lambda }=8.66$⁠.

$Z=-1.73$⁠. Since 60 is within 2 standard deviations of the mean, it would not generally be considered unusual. Note that we often use this rule of thumb even when the normal model does not apply.

Using Poisson with $\lambda =75$⁠: 0.0402.

$\frac{{\lambda }^k\times e^{-\lambda }}{k!}=$ $\frac{{6.5}^5\times e^{-6.5}}{5!}=$ $0.1454$

The probability will come to $0.0015+$ $0.0098+$ $0.0318=$ $0.0431$ (0.0430 if no rounding error).

The number of people per car is $11.7/6.5=1.8$⁠, meaning people are coming in small clusters. That is, if one person arrives, there’s a chance that they brought one or more other people in their vehicle. This means individuals (the people) are not independent, even if the car arrivals are independent, and this breaks a core assumption for the Poisson distribution. That is, the number of people visiting between 2pm and 3pm would not follow a Poisson distribution.

0 wins (-$3): 0.1458. 1 win (-$1): 0.3936. 2 wins (+$1): 0.3543. 3 wins (+$3): 0.1063.

Want to find the probability that there will be 1,787 or more enrollees. Using the normal approximation, with $\mu =np=\mathrm{2,500}\times 0.7=1750$ and $\sigma =$ $\sqrt{np(1-p)}=$ $\sqrt{\mathrm{2,500}\times 0.7\times 0.3}\approx$ $23$⁠, $Z=1.61$⁠, and $P(Z>1.61)=0.0537$⁠. With a 0.5 correction: 0.0559.

$Z=0.67$⁠.

$\mu =\$1650$⁠, $x=\$1800$⁠.

$0.67=$ $\frac{1800-1650}{\sigma }\to$ $\sigma =$ $\$223.88$⁠.

${(1-0.471)}^2\times 0.471=0.1318$⁠.

${0.471}^3=0.1045$⁠.

$\mu =1/0.471=2.12$⁠, $\sigma =\sqrt{2.38}=1.54$⁠.

$\mu =1/0.30=3.33$⁠, $\sigma =2.79$⁠.

When $p$ is smaller, the event is rarer, meaning the expected number of trials before a success and the standard deviation of the waiting time are higher.

$Z=1.56$⁠, $P(Z>1.56)=0.0594$⁠, i.e. 6%.

$Z=0.73$⁠, $P(Z>0.73)=0.2327$⁠.

If you are bidding on only one auction and set a low maximum bid price, someone will probably outbid you. If you set a high maximum bid price, you may win the auction but pay more than is necessary. If bidding on more than one auction, and you set your maximum bid price very low, you probably won’t win any of the auctions. However, if the maximum bid price is even modestly high, you are likely to win multiple auctions.

An answer roughly equal to the 10th percentile would be reasonable. Regrettably, no percentile cutoff point guarantees beyond any possible event that you win at least one auction. However, you may pick a higher percentile if you want to be more sure of winning an auction.

Answers will vary a little but should correspond to the answer in part (c). We use the 10⁠${}^{th}$ percentile: $Z=-1.28\to \$69.80$⁠.

$Z=3.5$⁠, upper tail is 0.0002. (More precise value: 0.000233, but we’ll use 0.0002 for the calculations here.)

$0.0002\times 2000=0.4$⁠. We would expect about 0.4 10 year olds who are 76 inches or taller to show up.

$\left(\frac{2000}{0}\right){(0.0002)}^0{(1-0.0002)}^{2000}=$ $0.67029$⁠.

$\frac{{0.4}^0\times e^{-0.4}}{0!}=$ $\frac{1\times e^{-0.4}}{1}=$ $0.67032$⁠.