QUIZ 3 (CHAPTER 6) - SOLUTIONS
QUIZ 3 (CHAPTER 6) - SOLUTIONS MATH 119 – FALL 2008 – KUNIYUKI 105 POINTS TOTAL, BUT 100 POINTS = 100% Show all work, simplify as appropriate, and use “good form and procedure” (as in class). Box in your final answers! No notes or books allowed. A scientific calculator is allowed.
Round off z scores to 2 decimal places; otherwise, write exact answers or round off calculations to at least four decimal places or at least four significant digits, whichever is more convenient. 1) (2 points). What is the mean of the standard normal distribution? 0
2) (2 points). What is the standard deviation of the standard normal distribution? 1
3) (2 points). Normal distributions are … (Box in one:) Continuous
Discrete
4) (2 points). We are drawing random samples from a distribution, D. Which of the following statements is true about the Central Limit Theorem (CLT)? Box in one:
i.
We can never use the CLT to study sample means if the D distribution is not normal.
ii.
We can sometimes use the CLT to study sample means, even if the D distribution is not normal. We can use the CLT if the sample size is large; in particular, if n > 30 .
5) (38 points total). The incomes of Oompa Loompas last year were approximately normally distributed with a mean of $5000 and a standard deviation of $1250. Do not use continuity corrections for these problems. a) What percent of Oompa Loompas made more than $8000 last year? Write your answer to the nearest hundredth of a percent. (10 points) Let X be the income (in $) last year of a randomly selected Oompa Loompa. If x = 8000, then: z =
(
)
x μ 8000 5000 = = 2.40 1250
(
)
Therefore, P X > 8000 P Z > 2.40 .
A 1 .9918 A .0082 A 0.82%
b) What percent of Oompa Loompas made between $6200 and $7600 last year? Write your answer to the nearest hundredth of a percent. (16 points) If x = 6200, then: z =
x μ 6200 5000 = = 0.96 1250
If x = 7600, then: z =
x μ 7600 5000 = = 2.08 1250
(
)
(
)
Therefore, P 6200 < X < 7600 P 0.96 < Z < 2.08 .
A .9812 .8315 A .1497 A 14.97%
c) The top 20% of Oompa Loompa income earners from last year are called “rich” in the Oompa Loompa community. The bottom 80% are called “poor.” Find the income level (from last year) that separates the “rich” Oompa Loompas from the “poor” Oompa Loompas. Write units! (12 points) Look in the body of Table A-2 to find the probability (or area) closest to 0.8000. The closest probability is 0.7995, and the corresponding z score is 0.84.
Find the corresponding x score: x = μ + z
(
)(
5000 + 0.84 1250 $6050
)
6) (36 points total). Dr. Evil has a thousand students in his lasers class. The scores on the final exam are approximately normally distributed with a mean of 23.5 points and a standard deviation of 2.8 points. Do not use continuity corrections for these problems. a) The students who scored between 23.0 points and 26.0 points on the final (and only those students) received a “C” grade on the final. What is the probability that a randomly selected student in the class received a “C” on the final? (16 points) Let X be the score (in points) of a randomly selected student in the class. If x = 23.0, then: z =
x μ 23.0 23.5 = 0.18 2.8
If x = 26.0, then: z =
x μ 26.0 23.5 = 0.89 2.8
(
)
(
)
Therefore, P 23.0 < X < 26.0 P 0.18 < Z < 0.89 .
A .8133 .4286 A .3847
b) We randomly select 11 students in the class. What is the probability that their average score on the final is between 23.0 points and 26.0 points? (20 points) Let X be the mean score (in points) of a group of 11 randomly selected students in the class. We can apply the Central Limit Theorem (CLT), because the original distribution is approximately normal. 2.8 X ~ N μ X = μ = 23.5, X = = 0.8442 11 n approx.
If x = 23.0, then: z = If x = 26.0, then: z =
(
)
(
x μX
X x μX
X
23.0 23.5 0.59 0.8442
26.0 23.5 2.96 0.8442
)
Therefore, P 23.0 < X < 26.0 P 0.59 < Z < 2.96 .
A .9985 .2776 A .7209
7) (23 points). In a special congressional Republican primary in Ohio, “frontrunner” Pat DeWine was humiliated when he came in fourth place with just 12% of the vote. A political think tank randomly selects 225 voters in that primary. Find the probability that fewer than (or “less than”) 20 of the 225 selected voters voted for DeWine in that primary. Use a normal approximation to a binomial distribution, and use a continuity correction. Show why the normal approximation is appropriate based on the rules given in class. Define “a success on one trial” as “the voter representing the trial voted for DeWine.” Then, p = 0.12, and q = 1 – p = 0.88. If we define X as the number of the selected voters who voted for DeWine, then X ~ Bin n = 225, p = 0.12 .
(
)
Show why the normal approximation is appropriate:
( )( ) (Good!) nq = ( 225) ( 0.88) = 198 5 ( Good!) np = 225 0.12 = 27 5
Therefore, we can use the approximation:
(
X ~ N μ = np = 27, = npq = approx.
( 225)(0.12)(0.88) 4.874)
Apply the continuity correction and go to z scores: The whole number “20” from the binomial distribution corresponds to the real numbers from 19.5 to 20.5 from the normal distribution. Since “fewer than” excludes 20, itself, we include only the numbers to the left of (i.e., lesser than) 19.5.
(
)
(
)
Therefore, P X < 20 P X C < 19.5 . If xC = 19.5, then: z
(
)
(
19.5 27 1.54 4.874
)
Therefore, P X < 20 P Z < 1.54 .
Find the desired probability:
A .0618
Considering that 20 out of 225 is about 8.89%, which may not seem that far from 12%, this probability may seem surprisingly low to some people. In Chapter 7, you may develop a better sense of probabilities involved with polls.
Round off z scores to 2 decimal places; otherwise, write exact answers or round off calculations to at least four decimal places or at least four significant digits, whichever is more convenient. 1) (2 points). What is the mean of the standard normal distribution? 0
2) (2 points). What is the standard deviation of the standard normal distribution? 1
3) (2 points). Normal distributions are … (Box in one:) Continuous
Discrete
4) (2 points). We are drawing random samples from a distribution, D. Which of the following statements is true about the Central Limit Theorem (CLT)? Box in one:
i.
We can never use the CLT to study sample means if the D distribution is not normal.
ii.
We can sometimes use the CLT to study sample means, even if the D distribution is not normal. We can use the CLT if the sample size is large; in particular, if n > 30 .
5) (38 points total). The incomes of Oompa Loompas last year were approximately normally distributed with a mean of $5000 and a standard deviation of $1250. Do not use continuity corrections for these problems. a) What percent of Oompa Loompas made more than $8000 last year? Write your answer to the nearest hundredth of a percent. (10 points) Let X be the income (in $) last year of a randomly selected Oompa Loompa. If x = 8000, then: z =
(
)
x μ 8000 5000 = = 2.40 1250
(
)
Therefore, P X > 8000 P Z > 2.40 .
A 1 .9918 A .0082 A 0.82%
b) What percent of Oompa Loompas made between $6200 and $7600 last year? Write your answer to the nearest hundredth of a percent. (16 points) If x = 6200, then: z =
x μ 6200 5000 = = 0.96 1250
If x = 7600, then: z =
x μ 7600 5000 = = 2.08 1250
(
)
(
)
Therefore, P 6200 < X < 7600 P 0.96 < Z < 2.08 .
A .9812 .8315 A .1497 A 14.97%
c) The top 20% of Oompa Loompa income earners from last year are called “rich” in the Oompa Loompa community. The bottom 80% are called “poor.” Find the income level (from last year) that separates the “rich” Oompa Loompas from the “poor” Oompa Loompas. Write units! (12 points) Look in the body of Table A-2 to find the probability (or area) closest to 0.8000. The closest probability is 0.7995, and the corresponding z score is 0.84.
Find the corresponding x score: x = μ + z
(
)(
5000 + 0.84 1250 $6050
)
6) (36 points total). Dr. Evil has a thousand students in his lasers class. The scores on the final exam are approximately normally distributed with a mean of 23.5 points and a standard deviation of 2.8 points. Do not use continuity corrections for these problems. a) The students who scored between 23.0 points and 26.0 points on the final (and only those students) received a “C” grade on the final. What is the probability that a randomly selected student in the class received a “C” on the final? (16 points) Let X be the score (in points) of a randomly selected student in the class. If x = 23.0, then: z =
x μ 23.0 23.5 = 0.18 2.8
If x = 26.0, then: z =
x μ 26.0 23.5 = 0.89 2.8
(
)
(
)
Therefore, P 23.0 < X < 26.0 P 0.18 < Z < 0.89 .
A .8133 .4286 A .3847
b) We randomly select 11 students in the class. What is the probability that their average score on the final is between 23.0 points and 26.0 points? (20 points) Let X be the mean score (in points) of a group of 11 randomly selected students in the class. We can apply the Central Limit Theorem (CLT), because the original distribution is approximately normal. 2.8 X ~ N μ X = μ = 23.5, X = = 0.8442 11 n approx.
If x = 23.0, then: z = If x = 26.0, then: z =
(
)
(
x μX
X x μX
X
23.0 23.5 0.59 0.8442
26.0 23.5 2.96 0.8442
)
Therefore, P 23.0 < X < 26.0 P 0.59 < Z < 2.96 .
A .9985 .2776 A .7209
7) (23 points). In a special congressional Republican primary in Ohio, “frontrunner” Pat DeWine was humiliated when he came in fourth place with just 12% of the vote. A political think tank randomly selects 225 voters in that primary. Find the probability that fewer than (or “less than”) 20 of the 225 selected voters voted for DeWine in that primary. Use a normal approximation to a binomial distribution, and use a continuity correction. Show why the normal approximation is appropriate based on the rules given in class. Define “a success on one trial” as “the voter representing the trial voted for DeWine.” Then, p = 0.12, and q = 1 – p = 0.88. If we define X as the number of the selected voters who voted for DeWine, then X ~ Bin n = 225, p = 0.12 .
(
)
Show why the normal approximation is appropriate:
( )( ) (Good!) nq = ( 225) ( 0.88) = 198 5 ( Good!) np = 225 0.12 = 27 5
Therefore, we can use the approximation:
(
X ~ N μ = np = 27, = npq = approx.
( 225)(0.12)(0.88) 4.874)
Apply the continuity correction and go to z scores: The whole number “20” from the binomial distribution corresponds to the real numbers from 19.5 to 20.5 from the normal distribution. Since “fewer than” excludes 20, itself, we include only the numbers to the left of (i.e., lesser than) 19.5.
(
)
(
)
Therefore, P X < 20 P X C < 19.5 . If xC = 19.5, then: z
(
)
(
19.5 27 1.54 4.874
)
Therefore, P X < 20 P Z < 1.54 .
Find the desired probability:
A .0618
Considering that 20 out of 225 is about 8.89%, which may not seem that far from 12%, this probability may seem surprisingly low to some people. In Chapter 7, you may develop a better sense of probabilities involved with polls.
