## Section4.6Exercises

### SubsectionExercises

###### 1Area under the curve, I

What percent of a standard normal distribution $N(\mu=0, \sigma=1)$ is found in each region? Be sure to draw a graph.

1. $Z \lt -1.35$ Answer

8.85%. 2. $Z \gt 1.48$ Answer

6.94%. 3. $-0.4 \lt Z \lt 1.5$ Answer

58.86%. 4. $|Z| \gt 2$ Answer

4.56%. ###### 2Area under the curve, II

What percent of a standard normal distribution $N(\mu=0, \sigma=1)$ is found in each region? Be sure to draw a graph.

1. $Z \gt -1.13$

2. $Z \lt 0.18$

3. $Z \gt 8$

4. $|Z| \lt 0.5$

###### 3GRE scores, Part I

Sophia who took the Graduate Record Examination (GRE) scored 160 on the Verbal Reasoning section and 157 on the Quantitative Reasoning section. The mean score for Verbal Reasoning section for all test takers was 151 with a standard deviation of 7, and the mean score for the Quantitative Reasoning was 153 with a standard deviation of 7.67. Suppose that both distributions are nearly normal.

1. Write down the short-hand for these two normal distributions. Answer

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

2. What is Sophia's Z-score on the Verbal Reasoning section? On the Quantitative Reasoning section? Draw a standard normal distribution curve and mark these two Z-scores. Answer

$Z_{VR} = 1.29\text{,}$ $Z_{QR} = 0.52\text{.}$ 3. What do these Z-scores tell you? Answer

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.

4. Relative to others, which section did she do better on? Answer

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

5. Find her percentile scores for the two exams. Answer

$Perc_{VR} = 0.9007 \approx 90\text{,}$ $Perc_{QR} = 0.6990 \approx 70\text{.}$

6. What percent of the test takers did better than her on the Verbal Reasoning section? On the Quantitative Reasoning section? Answer

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

7. Explain why simply comparing raw scores from the two sections could lead to an incorrect conclusion as to which section a student did better on. Answer

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.

8. If the distributions of the scores on these exams are not nearly normal, would your answers to parts (b) - (f) change? Explain your reasoning. Answer

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.

###### 4Triathlon times, Part I

In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups:

• The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds.

• The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds.

• The distributions of finishing times for both groups are approximately Normal.

Remember: a better performance corresponds to a faster finish.

1. Write down the short-hand for these two normal distributions.

2. What are the Z-scores for Leo's and Mary's finishing times? What do these Z-scores tell you?

3. Did Leo or Mary rank better in their respective groups? Explain your reasoning.

4. What percent of the triathletes did Leo finish faster than in his group?

5. What percent of the triathletes did Mary finish faster than in her group?

6. If the distributions of finishing times are not nearly normal, would your answers to parts (b) - (e) change? Explain your reasoning.

###### 5GRE scores, Part II

In Exercise 4.6.3 we saw two distributions for GRE scores: $N(\mu=151, \sigma=7)$ for the verbal part of the exam and $N(\mu=153, \sigma=7.67)$ for the quantitative part. Use this information to compute each of the following:

1. The score of a student who scored in the $80^{th}$ percentile on the Quantitative Reasoning section. Answer

$Z = 0.84\text{,}$ which corresponds to approximately 160 on QR.

2. The score of a student who scored worse than 70% of the test takers in the Verbal Reasoning section. Answer

$Z = -0.52\text{,}$ which corresponds to approximately 147 on VR.

###### 6Triathlon times, Part II

In Exercise 4.6.4 we saw two distributions for triathlon times: $N(\mu=4313, \sigma=583)$ for Men, Ages 30 - 34 and $N(\mu=5261, \sigma=807)$ for the Women, Ages 25 - 29 group. Times are listed in seconds. Use this information to compute each of the following:

1. The cutoff time for the fastest 5% of athletes in the men's group, i.e. those who took the shortest 5% of time to finish.

2. The cutoff time for the slowest 10% of athletes in the women's group.

###### 7LA weather, Part I

The average daily high temperature in June in LA is 77°F with a standard deviation of 5 °F. Suppose that the temperatures in June closely follow a normal distribution.

1. What is the probability of observing an 83°F temperature or higher in LA during a randomly chosen day in June? Answer

$Z=1.2 \to 0.1151\text{.}$

2. How cool are the coldest 10% of the days (days with lowest average high temperature) during June in LA? Answer

$Z= -1.28 \to$ 70.6°F or colder.

###### 8Portfolio returns

The Capital Asset Pricing Model is a financial model that assumes returns on a portfolio are normally distributed. Suppose a portfolio has an average annual return of 14.7% (i.e. an average gain of 14.7%) with a standard deviation of 33%. A return of 0% means the value of the portfolio doesn't change, a negative return means that the portfolio loses money, and a positive return means that the portfolio gains money.

1. What percent of years does this portfolio lose money, i.e. have a return less than 0%?

2. What is the cutoff for the highest 15% of annual returns with this portfolio?

###### 9LA weather, Part II

Exercise 4.6.7 states that average daily high temperature in June in LA is 77°F with a standard deviation of 5 °F, and it can be assumed that they to follow a normal distribution. We use the following equation to convert  °F (Fahrenheit) to  °C (Celsius):

\begin{equation*} C = (F - 32) \times \frac{5}{9}. \end{equation*}
1. Write the probability model for the distribution of temperature in °C in June in LA. Answer

$N(25, 2.78)\text{.}$

2. What is the probability of observing a 28°C (which roughly corresponds to 83 °F) temperature or higher in June in LA? Calculate using the  °C model from part (a). Answer

$Z = 1.08 \to 0.1401\text{.}$

3. Did you get the same answer or different answers in part (b) of this question and part (a) of Exercise 4.6.7? Are you surprised? Explain. Answer

The answers are very close because only the units were changed. (The only reason why they are a little different is because 28°C is 82.4 °F, not precisely 83 °F.)

4. Estimate the IQR of the temperatures (in °C) in June in LA. Answer

Since $IQR = Q3 - Q1\text{,}$ 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.

###### 10Heights of 10 year olds

Heights of 10 year olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches.

1. What is the probability that a randomly chosen 10 year old is shorter than 48 inches?

2. What is the probability that a randomly chosen 10 year old is between 60 and 65 inches?

3. If the tallest 10% of the class is considered “very tall”, what is the height cutoff for “very tall”?

4. The height requirement for Batman the Ride at Six Flags Magic Mountain is 54 inches. What percent of 10 year olds cannot go on this ride?

Suppose a newspaper article states that the distribution of auto insurance premiums for residents of California is approximately normal with a mean of $1,650. The article also states that 25% of California residents pay more than$1,800.

1. What is the Z-score that corresponds to the top 25% (or the $75^{th}$ percentile) of the standard normal distribution? Answer

$Z=0.67\text{.}$

2. What is the mean insurance cost? What is the cutoff for the 75th percentile? Answer

$\mu=1650\text{,}$ $x=1800\text{.}$

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

###### 12Speeding on the I-5, Part I

The distribution of passenger vehicle speeds traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. 1 S. Johnson and D. Murray. Empirical Analysis of Truck and Automobile Speeds on Rural Interstates: Impact of Posted Speed Limits. In: Transportation Research Board 89th Annual Meeting. 2010.

1. What percent of passenger vehicles travel slower than 80 miles/hour?

2. What percent of passenger vehicles travel between 60 and 80 miles/hour?

3. How fast do the fastest 5% of passenger vehicles travel?

4. The speed limit on this stretch of the I-5 is 70 miles/hour. Approximate what percentage of the passenger vehicles travel above the speed limit on this stretch of the I-5.

###### 13Overweight baggage, Part I

Suppose weights of the checked baggage of airline passengers follow a nearly normal distribution with mean 45 pounds and standard deviation 3.2 pounds. Most airlines charge a fee for baggage that weigh in excess of 50 pounds. Determine what percent of airline passengers incur this fee.

$Z = 1.56 \to 0.0594\text{,}$ i.e. 6%.

###### 14Find the SD

Find the standard deviation of the distribution in the following situations.

1. MENSA is an organization whose members have IQs in the top 2% of the population. IQs are normally distributed with mean 100, and the minimum IQ score required for admission to MENSA is 132.

2. Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter (mg/dl). Women with cholesterol levels above 220 mg/dl are considered to have high cholesterol and about 18.5% of women fall into this category.

The textbook you need to buy for your chemistry class is expensive at the college bookstore, so you consider buying it on Ebay instead. A look at past auctions suggest that the prices of that chemistry textbook have an approximately normal distribution with mean $89 and standard deviation$15.

###### 25Ages of pennies, Part I

The histogram below shows the distribution of ages of pennies at a bank.

1. Describe the distribution. The distribution is unimodal and strongly right skewed with a median between 5 and 10 years old. Ages range from 0 to slightly over 50 years old, and the middle 50% of the distribution is roughly between 5 and 15 years old. There are potential outliers on the higher end.

2. Sampling distributions for means from simple random samples of 5, 30, and 100 pennies is shown in the histograms below. Describe the shapes of these distributions and comment on whether they look like what you would expect to see based on the Central Limit Theorem.   When the sample size is small, the sampling distribution is right skewed, just like the population distribution. As the sample size increases, the sampling distribution gets more unimodal, symmetric, and approaches normality. The variability also decreases. This is consistent with the Central Limit Theorem.

###### 26Ages of pennies, Part II

The mean age of the pennies from Exercise 4.6.25 is 10.44 years with a standard deviation of 9.2 years. Using the Central Limit Theorem, calculate the means and standard deviations of the distribution of the mean from random samples of size 5, 30, and 100. Comment on whether the sampling distributions shown in Exercise 4.6.25 agree with the values you compute.

###### 27Housing prices, Part I

A housing survey was conducted to determine the price of a typical home in Topanga, CA. The mean price of a house was roughly $1.3 million with a standard deviation of$300,000. There were no houses listed below $600,000 but a few houses above$3 million.

1. Is the distribution of housing prices in Topanga symmetric, right skewed, or left skewed? Hint: Sketch the distribution. Answer

Right skewed. There is a long tail on the higher end of the distribution but a much shorter tail on the lower end.

2. Would you expect most houses in Topanga to cost more or less than $1.3 million? Answer Less than, as the median would be less than the mean in a right skewed distribution. 3. Can we estimate the probability that a randomly chosen house in Topanga costs more than$1.4 million using the normal distribution? Answer

We should not.

###### 39Is it Bernoulli?

Determine if each trial can be considered an independent Bernoulli trial for the following situations.

1. Cards dealt in a hand of poker. Answer

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.

2. Outcome of each roll of a die. Answer

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 simply to two events, e.g. rolling a 6 and not rolling a 6, though specifying such details would be necessary.

###### 40With and without replacement

In the following situations assume that half of the specified population is male and the other half is female.

1. Suppose you're sampling from a room with 10 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement?

2. Now suppose you're sampling from a stadium with 10,000 people. What is the probability of sampling two females in a row when sampling with replacement? What is the probability when sampling without replacement?

3. We often treat individuals who are sampled from a large population as independent. Using your findings from parts (a) and (b), explain whether or not this assumption is reasonable.

###### 41Married women

The 2010 American Community Survey estimates that 47.1% of women ages 15 years and over are married. 2 U.S. Census Bureau, 2010 American Community Survey, Marital Status.

1. We randomly select three women between these ages. What is the probability that the third woman selected is the only one who is married? Answer

$(1-0.471)^2\times0.471 = 0.1318\text{.}$

2. What is the probability that all three randomly selected women are married? Answer

$0.471^3 = 0.1045\text{.}$

3. On average, how many women would you expect to sample before selecting a married woman? What is the standard deviation? Answer

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

4. If the proportion of married women was actually 30%, how many women would you expect to sample before selecting a married woman? What is the standard deviation? Answer

$\mu = 1/0.30 = 3.33\text{,}$ $\sigma=2.79\text{.}$

5. Based on your answers to parts (c) and (d), how does decreasing the probability of an event affect the mean and standard deviation of the wait time until success? Answer

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.

###### 42Defective rate

A machine that produces a special type of transistor (a component of computers) has a 2% defective rate. The production is considered a random process where each transistor is independent of the others.

1. What is the probability that the $10^{th}$ transistor produced is the first with a defect?

2. What is the probability that the machine produces no defective transistors in a batch of 100?

3. On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation?

4. Another machine that also produces transistors has a 5% defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation?

5. Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?

###### 43Eye color, Part I

A husband and wife both have brown eyes but carry genes that make it possible for their children to have brown eyes (probability 0.75), blue eyes (0.125), or green eyes (0.125).

1. What is the probability the first blue-eyed child they have is their third child? Assume that the eye colors of the children are independent of each other. Answer

$0.875^2\times 0.125 = 0.096\text{.}$

2. On average, how many children would such a pair of parents have before having a blue-eyed child? What is the standard deviation of the number of children they would expect to have until the first blue-eyed child? Answer

$\mu=8\text{,}$ $\sigma=7.48\text{.}$

###### 44Speeding on the I-5, Part II

Exercise 4.6.12 states that the distribution of speeds of cars traveling on the Interstate 5 Freeway (I-5) in California is nearly normal with a mean of 72.6 miles/hour and a standard deviation of 4.78 miles/hour. The speed limit on this stretch of the I-5 is 70 miles/hour.

1. A highway patrol officer is hidden on the side of the freeway. What is the probability that 5 cars pass and none are speeding? Assume that the speeds of the cars are independent of each other.

2. On average, how many cars would the highway patrol officer expect to watch until the first car that is speeding? What is the standard deviation of the number of cars he would expect to watch?

###### 45Underage drinking, Part II

We learned in Exercise 3.7.35 that about 70% of 18-20 year olds consumed alcoholic beverages in 2008. We now consider a random sample of fifty 18-20 year olds.

1. How many people would you expect to have consumed alcoholic beverages? And with what standard deviation? Answer

$\mu=35\text{,}$ $\sigma=3.24\text{.}$

2. Would you be surprised if there were 45 or more people who have consumed alcoholic beverages? Answer

Yes. $Z=3.09\text{.}$ Since 45 is more than 2 standard deviations from the mean, it would be considered unusual. Note that the normal model is not required to apply this rule of thumb.

3. What is the probability that 45 or more people in this sample have consumed alcoholic beverages? How does this probability relate to your answer to part (b)? Answer

Using a normal model: 0.0010. This does indeed appear to be an unusual observation. If using a normal model with a 0.5 correction, the probability would be calculated as 0.0017.

###### 46Chickenpox, Part II

1. How many people in this sample would you expect to have had chickenpox in their childhood? And with what standard deviation?

2. Would you be surprised if there were 105 people who have had chickenpox in their childhood?

3. What is the probability that 105 or fewer people in this sample have had chickenpox in their childhood? How does this probability relate to your answer to part (b)?

Suppose a university announced that it admitted 2,500 students for the following year's freshman class. However, the university has dorm room spots for only 1,786 freshman students. If there is a 70% chance that an admitted student will decide to accept the offer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class?

Want to find the probability that there will be 1,786 or more enrollees. Using the normal: 0.0582. With a 0.5 correction: 0.0559.

###### 48Survey response rate

Pew Research reported in 2012 that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond? 3 The Pew Research Center for the People and the Press, Assessing the Representativeness of Public Opinion Surveys, May 15, 2012.

###### 49Game of dreidel

A dreidel is a four-sided spinning top with the Hebrew letters nun, gimel, hei, and shin, one on each side. Each side is equally likely to come up in a single spin of the dreidel. Suppose you spin a dreidel three times. Calculate the probability of getting

1. at least one nun? Answer

$1-0.75^3 = 0.5781\text{.}$

0.1406.

0.4219.

4. at most 2 gimels? Answer

$1-0.25^3=0.9844\text{.}$ ###### 50Arachnophobia

A 2005 Gallup Poll found that 7% of teenagers (ages 13 to 17) suffer from arachnophobia and are extremely afraid of spiders. At a summer camp there are 10 teenagers sleeping in each tent. Assume that these 10 teenagers are independent of each other. 4 Gallup Poll, What Frightens America's Youth?, March 29, 2005.

1. Calculate the probability that at least one of them suffers from arachnophobia.

2. Calculate the probability that exactly 2 of them suffer from arachnophobia.

3. Calculate the probability that at most 1 of them suffers from arachnophobia.

4. If the camp counselor wants to make sure no more than 1 teenager in each tent is afraid of spiders, does it seem reasonable for him to randomly assign teenagers to tents?

###### 51Eye color, Part II

Exercise 4.6.43 introduces a husband and wife with brown eyes who have 0.75 probability of having children with brown eyes, 0.125 probability of having children with blue eyes, and 0.125 probability of having children with green eyes.

1. What is the probability that their first child will have green eyes and the second will not? Answer

Geometric distribution: 0.109.

2. What is the probability that exactly one of their two children will have green eyes? Answer

Binomial: 0.219.

3. If they have six children, what is the probability that exactly two will have green eyes? Answer

Binomial: 0.137.

4. If they have six children, what is the probability that at least one will have green eyes? Answer

$1-0.875^6=0.551\text{.}$

5. What is the probability that the first green eyed child will be the $4^{th}$ child? Answer

Geometric: 0.084.

6. Would it be considered unusual if only 2 out of their 6 children had brown eyes? Answer

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

###### 52Sickle cell anemia

Sickle cell anemia is a genetic blood disorder where red blood cells lose their flexibility and assume an abnormal, rigid, “sickle” shape, which results in a risk of various complications. If both parents are carriers of the disease, then a child has a 25% chance of having the disease, 50% chance of being a carrier, and 25% chance of neither having the disease nor being a carrier. If two parents who are carriers of the disease have 3 children, what is the probability that

1. two will have the disease?

2. none will have the disease?

3. at least one will neither have the disease nor be a carrier?

4. the first child with the disease will the be $3^{rd}$ child?

###### 53Roulette winnings

In the game of roulette, a wheel is spun and you place bets on where it will stop. One popular bet is that it will stop on a red slot; such a bet has an 18/38 chance of winning. If it stops on red, you double the money you bet. If not, you lose the money you bet. Suppose you play 3 times, each time with a $1 bet. Let Y represent the total amount won or lost. Write a probability model for Y. Answer 0 wins (-$3): 0.1458. 1 win (-$1): 0.3936. 2 wins (+$1): 0.3543. 3 wins (+\$3): 0.1063.

###### 54Multiple choice quiz

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that

1. the first question she gets right is the $3^{rd}$ question?

2. she gets exactly 3 or exactly 4 questions right?

3. she gets the majority of the questions right?

###### 55Distribution of $\hat{p}$

Suppose the true population proportion were $p = 0.95\text{.}$ The figure below shows what the distribution of a sample proportion looks like when the sample size is $n = 20\text{,}$ $n = 100\text{,}$ and $n = 500\text{.}$ (a) What does each point (observation) in each of the samples represent? (b) Describe the distribution of the sample proportion, $\hat{p}\text{.}$ How does the distribution of the sample proportion change as $n$ becomes larger? (a) Each observation in each of the distributions represents the sample proportion ($\hat{p}$) from samples of size $n = 20\text{,}$ $n = 100\text{,}$ and $n = 500\text{,}$ respectively. (b) The centers for all three distributions are at 0.95, the true population parameter. When $n$ is small, the distribution is skewed to the left and not smooth. As $n$ increases, the variability of the distribution (standard deviation) decreases, and the shape of the distribution becomes more unimodal and symmetric.

###### 56Distribution of $\hat{p}$

Suppose the true population proportion were $p = 0.5\text{.}$ The figure below shows what the distribution of a sample proportion looks like when the sample size is $n = 20\text{,}$ $n = 100\text{,}$ and $n = 500\text{.}$ What does each point (observation) in each of the samples represent? Describe how the distribution of the sample proportion, $\hat{p}\text{,}$ changes as $n$ becomes larger. ###### 57Distribution of $\hat{p}$

Suppose the true population proportion were $p = 0.5$ and a researcher takes a simple random sample of size $n=50\text{.}$

1. Find and interpret the standard deviation of the sample proportion $\hat{p}\text{.}$ Answer

$SD_{\hat{p}} = \sqrt{p(1-p) / n} = 0.0707\text{.}$ This describes the typical distance that the sample proportion will deviate from the true proportion, $p = 0.5\text{.}$

2. Calculate the probability that the sample proportion will be larger than 0.55 for a random sample of size 50. Answer

$\hat{p}$ approximately follows $N(0.5, 0.0707)\text{.}$ $Z = (0.55 - 0.50) / 0.0707 \approx 0.71\text{.}$ This corresponds to an upper tail of about 0.2389. That is, $P(\hat{p} \gt 0.55) \approx 0.24\text{.}$

###### 58Distribution of $\hat{p}$

Suppose the true population proportion were $p = 0.6$ and a researcher takes a simple random sample of size $n=50\text{.}$

1. Find and interpret the standard deviation of the sample proportion $\hat{p}\text{.}$

2. Calculate the probability that the sample proportion will be larger than 0.65 for a random sample of size 50.

###### 59Nearsighted children

It is believed that nearsightedness affects about 8% of all children. We are interested in finding the probability that fewer than 12 out of 200 randomly sampled children will be nearsighted.

1. Estimate this probability using the normal approximation to the binomial distribution. Answer

First we need to check that the necessary conditions are met. There are $200 \times 0.08 = 16$ expected successes and $200 \times (1 - 0.08) = 184$ expected failures, therefore the success-failure condition is met. Then the binomial distribution can be approximated by $N(\mu = 16, \sigma = 3.84)\text{.}$ $P(X \lt 12) = P(Z \lt -1.04) = 0.1492\text{.}$

2. Estimate this probability using the distribution of the sample proportion. Answer

Since the success-failure condition is met the sampling distribution of $\hat{p} \sim N(\mu = 0.08, \sigma = 0.0192)\text{.}$ $P(\hat{p} \lt 0.06) = P(Z \lt -1.04) = 0.1492\text{.}$

As expected, the two answers are the same.

###### 60Poverty in the US

The 2013 Current Population Survey (CPS) estimates that 22.5% of Mississippians live in poverty, which makes Mississippi the state with the highest poverty rate in the United States. 5 United States Census Bureau. 2013 Current Population Survey. Historical Poverty Tables - People.Web. We are interested in finding out the probability that at least 250 people among a random sample of 1,000 Mississippians live in poverty.

1. Estimate this probability using the normal approximation to the binomial distribution.

2. Estimate this probability using the distribution of the sample proportion.

First we need to check that the necessary conditions are met. There are $100 \times 0.389 = 38.9$ expected successes and $100 \times (1 - 0.389) = 61.1$ expected failures, therefore the success-failure condition is met. Calculate using either (1) the normal approximation to the binomial distribution or (2) the sampling distribution of $\hat{p}\text{.}$ (1) The binomial distribution can be approximated by $N(\mu = 0.389, \sigma = 4.88)\text{.}$ $P(X \ge 35) = P(Z \gt -0.80) = 1 - 0.2119 = 0.7881\text{.}$ (2) The sampling distribution of $\hat{p} \sim N(\mu = 0.389, \sigma = 0.0488)\text{.}$ $P(\hat{p} \gt 0.35) = P(Z \gt -0.8) = 0.7881\text{.}$