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# Central Limit Theorem: Homework

Summary: Note: This module is currently under revision, and its content is subject to change. This module is being prepared as part of a statistics textbook that will be available for the Fall 2008 semester.

Note: You are viewing an old version of this document. The latest version is available here.

## Exercise 1

X~N(60,9)X~N(60,9) size 12{X "~" N $$"60",9$$ } {}. Suppose that you form random samples of 25 from this distribution. Let X¯X¯ size 12{ {overline {X}} } {} be the random variable of averages. Let ΣXΣX size 12{ΣX} {} be the random variable of sums. For c - f, sketch the graph, shade the region, label and scale the horizontal axis for X¯X¯ size 12{ {overline {X}} } {}, and find the probability.

• a: Sketch the distributions of XX size 12{X} {} and X¯X¯ size 12{ {overline {X}} } {} on the same graph.
• b: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {}
• c: P ( X ¯ < 60 ) = P ( X ¯ < 60 ) = size 12{P $${overline {X}} <"60"$$ ={}} {}
• d: Find the 30th percentile.
• e: P ( 56 < X ¯ < 62 ) = P ( 56 < X ¯ < 62 ) = size 12{P $$"56"< {overline {X}} <"62"$$ ={}} {}
• f: P ( 18 < X ¯ < 58 ) = P ( 18 < X ¯ < 58 ) = size 12{P $$"18"< {overline {X}} <"58"$$ ={}} {}
• g: Σ Σ size 12{Σ} {}
• h: Find the minimum value for the upper quartile.
• i: P ( 1400 < ΣX < 1550 ) = P ( 1400 < ΣX < 1550 ) = size 12{P $$"1400"<ΣX<"1550"$$ ={}} {}

### Solution

X ~ N ( 60 , 9 ) X ~ N ( 60 , 9 ) size 12{X "~" N $$"60",9$$ } {} mean = μ = 60 = μ = 60 size 12{ {}=μ="60"} {} , standard deviation = σ = 9 = σ = 9 size 12{ {}=σ=9} {} , sample size = n = 25 = n = 25 size 12{ {}=n="25"} {} . Random Variables: X ¯ X ¯ size 12{ {overline {X}} } {} is equal to Xbar. ΣX ΣX size 12{ΣX} {} is equal to sumX.

Note: z z size 12{z - {}} {} score formula for Xbar Xbar size 12{ ital "Xbar"} {} : z = ( xbar μ ) ( σ n ) z = ( xbar μ ) ( σ n ) size 12{z= { { $$ital "xbar" - μ$$ } over { $${ {σ} over { sqrt {n} } }$$ } } } {} ; z z size 12{z - {}} {} score formula for SumX SumX size 12{ ital "SumX"} {} : z = ( Σx ) ( n σ ) z = ( Σx ) ( n σ ) size 12{z= { { $$Σx - nμ$$ } over { $$sqrt {n} * σ$$ } } } {}

• a.:
• b.: Xbar ~ N ( 60 , 9 25 ) Xbar ~ N ( 60 , 9 25 ) size 12{ ital "Xbar" "~" N $$"60", { {9} over { sqrt {"25"} } }$$ } {}
• c.: P ( Xbar < 60 ) = P ( Z < 0 ) = 0 . 5 P ( Xbar < 60 ) = P ( Z < 0 ) = 0 . 5 size 12{P $$ital "Xbar"<"60"$$ =P $$Z<0$$ =0 "." 5} {}
• d.: P ( Xbar < k ) = 0 . 30 P ( Xbar < k ) = 0 . 30 size 12{P $$ital "Xbar"<k$$ =0 "." "30"} {} k = 59 . 06 k = 59 . 06 size 12{k="59" "." "06"} {} (Note: need graph) ( z = 0 . 5244 ) ( z = 0 . 5244 ) size 12{ $$z= - 0 "." "5244"$$ } {}
• e.: P ( 56 < Xbar < 62 ) = P ( 2 . 2222 < Z < 1 . 1111 ) = 0 . 8667 0 . 0131 = 0 . 8536 P ( 56 < Xbar < 62 ) = P ( 2 . 2222 < Z < 1 . 1111 ) = 0 . 8667 0 . 0131 = 0 . 8536 size 12{P $$"56"< ital "Xbar"<"62"$$ =P $$- 2 "." "2222"<Z<1 "." "1111"$$ =0 "." "8667" - 0 "." "0131"=0 "." "8536"} {}
• f.: P ( 18 < Xbar < 58 ) = P ( 23 . 3333 < Z < 1 . 1111 ) = 0 . 1333 0 = 0 . 1333 P ( 18 < Xbar < 58 ) = P ( 23 . 3333 < Z < 1 . 1111 ) = 0 . 1333 0 = 0 . 1333 size 12{P $$"18"< ital "Xbar"<"58"$$ =P $$- "23" "." "3333"<Z< - 1 "." "1111"$$ =0 "." "1333" - 0=0 "." "1333"} {}
• g.: SumX ~ N ( 25 60 , 25 9 ) SumX ~ N ( 25 60 , 25 9 ) size 12{ ital "SumX" "~" N $$"25" * "60", sqrt {"25"} * 9$$ } {}
• h.: P(ΣX>k)=0.25P(ΣX>k)=0.25 size 12{P $$ΣX>k$$ =0 "." "25"} {}P(ΣX>k)=0.75P(ΣX>k)=0.75 size 12{P $$ΣX>k$$ =0 "." "75"} {}; (z=0.06745)(z=0.06745) size 12{ $$z=0 "." "06745"$$ } {}k=1530.35k=1530.35 size 12{k="1530" "." "35"} {}
• i.: P ( 1400 < SumX < 1550 ) = P ( 2 . 2222 < Z < 1 . 1111 ) = 0 . 8667 0 . 0131 = 0 . 8536 P ( 1400 < SumX < 1550 ) = P ( 2 . 2222 < Z < 1 . 1111 ) = 0 . 8667 0 . 0131 = 0 . 8536 size 12{P $$"1400"< ital "SumX"<"1550"$$ =P $$- 2 "." "2222"<Z<1 "." "1111"$$ =0 "." "8667" - 0 "." "0131"=0 "." "8536"} {}

## Exercise 2

Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.

• a: When the sample size is large, the mean of X¯X¯ size 12{ {overline {X}} } {} is approximately equal to the mean of XX size 12{X} {}.
• b: When the sample size is large, X¯X¯ size 12{ {overline {X}} } {} is approximately normally distributed.
• c: When the sample size is large, the standard deviation of X¯X¯ size 12{ {overline {X}} } {} is approximately the same as the standard deviation of XX size 12{X} {}.

## Exercise 3

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about 10. Suppose that 16 individuals are randomly chosen.

Let X¯=X¯= size 12{ {overline {X}} ={}} {}average percent of fat calories.

• a: X¯~X¯~ size 12{ {overline {X}} "~" } {}______ ( ______ , ______ )
• b: For the group of 16, find the probability that the average percent of fat calories consumed is more than 5. Graph the situation and shade in the area to be determined.
• c: Find the first quartile for the average percent of fat calories.

### Solution

• a: Xbar~N(36,1016)Xbar~N(36,1016) size 12{ ital "Xbar" "~" N $$"36", { {"10"} over { sqrt {"16"} } }$$ } {} mean =36=36 size 12{ {}="36"} {} standard deviation =1016=104=2.5=1016=104=2.5 size 12{ {}= { {"10"} over { sqrt {"16"} } } = { {"10"} over {4} } =2 "." 5} {}
• b: P ( Xbar > 5 ) = 1 P ( Xbar > 5 ) = 1 size 12{P $$ital "Xbar">5$$ =1} {}
• c: Find kk size 12{k} {} such that P(Xbar<k)=0.25P(Xbar<k)=0.25 size 12{P $$ital "Xbar"<k$$ =0 "." "25"} {}; k=34.3138k=34.3138 size 12{k="34" "." "3138"} {}

## Exercise 4

Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students. • a: In words, X=X= size 12{X={}} {} • b: X ~ X ~ size 12{X "~" } {} • c: In words, X¯=X¯= size 12{ {overline {X}} ={}} {} • d: X¯~X¯~ size 12{ {overline {X}} "~" } {} ______ ( ______ , ______ ) • e: Find the probability that an individual had between$0.80 and $1.00. Graph the situation and shade in the area to be determined. • f: Find the probability that the average of the 25 students was between$0.80 and $1.00. Graph the situation and shade in the area to be determined. • g: Explain the why there is a difference in (e) and (f). ## Exercise 5 Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls. • a: If X¯=X¯= size 12{ {overline {X}} ={}} {} average distance in feet for 49 fly balls, then X¯~X¯~ size 12{ {overline {X}} "~" } {}____ ( _____ , _____ ) • b: What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for X¯X¯ size 12{ {overline {X}} } {}. Shade the region corresponding to the probability. Find the probability. • c: Find the 80th percentile of the distribution of the average of 49 fly balls. ### Solution • a: Xbar ~ N ( 250 , 50 49 ) Xbar ~ N ( 250 , 50 49 ) size 12{ ital "Xbar" "~" N $$"250", { {"50"} over { sqrt {"49"} } }$$ } {} • b: P ( Xbar < 240 ) = P ( Z < 1 . 4 ) = 0 . 0808 P ( Xbar < 240 ) = P ( Z < 1 . 4 ) = 0 . 0808 size 12{P $$ital "Xbar"<"240"$$ =P $$Z< - 1 "." 4$$ =0 "." "0808"} {} • c: Find kk size 12{k} {} such that P(Xbar<k)=0.80P(Xbar<k)=0.80 size 12{P $$ital "Xbar"<k$$ =0 "." "80"} {}; (z=0.8416)(z=0.8416) size 12{ $$z=0 "." "8416"$$ } {} k=256.0116k=256.0116 size 12{k="256" "." "0116"} {} ## Exercise 6 Suppose that the weight of open boxes of cereal in a home with children is uniformly distributed from 2 to 6 pounds. We randomly survey 64 homes with children. • a: In words, X=X= size 12{X={}} {} • b: X ~ X ~ size 12{X "~" } {} • c: μ X = μ X = size 12{μ rSub { size 8{X} } ={}} {} • d: σ X = σ X = size 12{σ rSub { size 8{X} } ={}} {} • e: In words, ΣX=ΣX= size 12{ΣX={}} {} • f: ΣX ~ ΣX ~ size 12{ΣX "~" } {} • g: Find the probability that the total weight of open boxes is less than 250 pounds. • h: Find the 35th percentile for the total weight of open boxes of cereal. ## Exercise 7 Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of 7 days. We randomly sample 9 trials. • a: In words, ΣX=ΣX= size 12{ΣX={}} {} • b: ΣX ~ ΣX ~ size 12{ΣX "~" } {} • c: Find the probability that the total length of the 9 trials is at least 225 days. • d: 90 percent of the total of 9 of these types of trials will last at least how long? ### Solution • a: SumX=SumX= size 12{ ital "SumX"={}} {} the total time 9 criminal trials last • b: SumX~N(921,97)SumX~N(921,97) size 12{ ital "SumX" "~" N $$9 * "21", sqrt {9} * 7$$ } {} mean =189=189 size 12{ {}="189"} {} standard deviation =21=21 size 12{ {}="21"} {} • c: P(SumX225)=P(Z>1.7143)=0.0432P(SumX225)=P(Z>1.7143)=0.0432 size 12{P $$ital "SumX" >= "225"$$ =P $$Z>1 "." "7143"$$ =0 "." "0432"} {}; z=((225nμ)nσ)z=((225nμ)nσ) size 12{z= $${ { \( "225" - n * μ$$ } over { sqrt {n} * σ} } \) } {} • d: Find kk size 12{k} {} such that P(SumXk)=0.90P(SumXk)=0.90 size 12{P $$ital "SumX" >= k$$ =0 "." "90"} {}; One way is the find the 10th %ile where P(SumX<k)=0.10P(SumX<k)=0.10 size 12{P $$ital "SumX"<k$$ =0 "." "10"} {}; k=162.0874k=162.0874 size 12{k="162" "." "0874"} {} ## Exercise 8 According to the Internal Revenue Service, the average length of time for an individual to complete (record keep, learn, prepare, copy, assemble and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is 2 hours. Suppose we randomly sample 36 taxpayers. • a: In words, X=X= size 12{X={}} {} • b: In words, X¯=X¯= size 12{ {overline {X}} ={}} {} • c: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • d: Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences. • e: Would you be surprised if one taxpayer finished his Form 1040 in more than 12 hours? In a complete sentence, explain why. ## Exercise 9 Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let X¯=X¯= size 12{ {overline {X}} ={}} {} the average of the 49 races. • a: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • b: Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons. • c: Find the 80th percentile for the average of these 49 marathons. • d: Find the median of the average running times. ### Solution • a: Xbar~N(145,1449)Xbar~N(145,1449) size 12{ ital "Xbar" "~" N $$"145", { {"14"} over { sqrt {"49"} } }$$ } {} mean =145=145 size 12{ {}="145"} {} standard deviation =2=2 size 12{ {}=2} {} • b: P ( 142 < Xbar < 146 ) = P ( 1 . 5 < Z < 0 . 5 ) = 0 . 6915 0 . 0668 = 0 . 6247 P ( 142 < Xbar < 146 ) = P ( 1 . 5 < Z < 0 . 5 ) = 0 . 6915 0 . 0668 = 0 . 6247 size 12{P $$"142"< ital "Xbar"<"146"$$ =P $$- 1 "." 5<Z<0 "." 5$$ =0 "." "6915" - 0 "." "0668"=0 "." "6247"} {} • c: Find kk size 12{k} {} such that P(Xbar<k)=0.80P(Xbar<k)=0.80 size 12{P $$ital "Xbar"<k$$ =0 "." "80"} {}; k=146.6832k=146.6832 size 12{k="146" "." "6832"} {} • d: In theory, the median is equal to the mean; median =145=145 size 12{ {}="145"} {} ## Exercise 10 The attention span of a two year-old is exponentially distributed with a mean of about 8 minutes. Suppose we randomly survey 60 two year-olds. • a: In words, X=X= size 12{X={}} {} • b: X ~ X ~ size 12{X "~" } {} • c: In words, X¯=X¯= size 12{ {overline {X}} ={}} {} • d: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • e: Before doing any calculations, which do you think will be higher? Explain why. • i: the probability that an individual attention span is less than 10 minutes; or • ii: the probability that the average attention span for the 60 children is less than 10 minutes? • f: Calculate the probabilities in part (e). • g: Explain why the distribution for X¯X¯ size 12{ {overline {X}} } {} is not exponential. ## Exercise 11 Suppose that the length of research papers is uniformly distributed from 10 to 25 pages. We survey a class in which 55 research papers were turned in to a professor. We are interested in the average length of the research papers. • a: In words, X=X= size 12{X={}} {} • b: X ~ X ~ size 12{X "~" } {} • c: μ X = μ X = size 12{μ rSub { size 8{X} } ={}} {} • d: σ X = σ X = size 12{σ rSub { size 8{X} } ={}} {} • e: In words, X¯=X¯= size 12{ {overline {X}} ={}} {} • f: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • g: In words, ΣX = ΣX = size 12{ΣX={}} {} • h: ΣX ~ ΣX ~ size 12{ΣX "~" } {} • i: Without doing any calculations, do you think that it’s likely that the professor will need to read a total of more than 1050 pages? Why? • j: Calculate the probability that the professor will need to read a total of more than 1050 pages. • k: Why is it so unlikely that the average length of the papers will be less than 12 pages? ### Solution • a: X=X= size 12{X={}} {}the length of a research paper, in pages • b: X~U(10,25)X~U(10,25) size 12{X "~" U $$"10","25"$$ } {} (Uniform) • c: Mean =(10+25)2=17.5=(10+25)2=17.5 size 12{ {}= { { $$"10"+"25"$$ } over {2} } ="17" "." 5} {} • d: Stdev =((2510)212=(22512)=((2510)212=(22512) size 12{ {}= sqrt { $${ { \( "25" - "10"$$ rSup { size 8{2} } } over {"12"} } } = sqrt { $${ {"225"} over {"12"} }$$ } } {} • e: Xbar=Xbar= size 12{ ital "Xbar"={}} {}the average length of 55 research papers, in pages • f: Xbar~N(17.5,2251255Xbar~N(17.5,2251255 size 12{ ital "Xbar" "~" N $$"17" "." 5, sqrt { left ( { { left ( { {"225"} over {"12"} } right )} over {"55"} } right )} } {} mean =17.5=17.5 size 12{ {}="17" "." 5} {} standard deviation =(1512)55=0.5839=(1512)55=0.5839 size 12{ {}= { { \( { {"15"} over { sqrt {"12"} } }$$ } over { sqrt {"55"} } } =0 "." "5839"} {} • g: SumX=SumX= size 12{ ital "SumX"={}} {}the total length of 55 research papers, in pages • h: SumX~N(5517.5,(55(22512)))SumX~N(5517.5,(55(22512))) size 12{ ital "SumX" "~" N $$"55" * "17" "." 5, sqrt { \( "55" * \( { {"225"} over {"12"} }$$ \) } \) } {} mean =962.5=962.5 size 12{ {}="962" "." 5} {} standard deviation =32.1131=32.1131 size 12{ {}="32" "." "1131"} {} • i: No. The value 1050 is more than 2 standard deviation above the mean. • j: P ( SumX > 1050 ) = P ( Z > 2 . 7247 ) = 0 . 0032 P ( SumX > 1050 ) = P ( Z > 2 . 7247 ) = 0 . 0032 size 12{P $$ital "SumX">"1050"$$ =P $$Z>2 "." "7247"$$ =0 "." "0032"} {} • k.: P(Xbar<12)=P(Z<9.4198)=0P(Xbar<12)=P(Z<9.4198)=0 size 12{P $$ital "Xbar"<"12"$$ =P $$Z< - 9 "." "4198"$$ =0} {}. It is unlikely that the average length of the papers is less than 12 because the probability is almost 0. 12 is more than 9 stdev below the ## Exercise 12 The length of songs in a collector’s CD collection is uniformly distributed from 2 to 3.5 minutes. Suppose we randomly pick 5 CDs from the collection. There is a total of 43 songs on the 5 CDs. • a: In words, X=X= size 12{X={}} {} • b: X ~ X ~ size 12{X "~" } {} • c: In words, X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • d: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • e: Find the first quartile for the average song length. • f: The IQR (interquartile range) for the average song length is from __________ to __________. ## Exercise 13 Salaries for teachers in a particular elementary school district are normally distributed with a mean of$44,000 and a standard deviation of $6500. We randomly survey 10 teachers from that district. • a: In words, X=X= size 12{X={}} {} • b: In words, X¯=X¯= size 12{ {overline {X}} ={}} {} • c: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {} • d: In words, ΣX=ΣX= size 12{ΣX={}} {} • e: ΣX ~ ΣX ~ size 12{ΣX "~" } {} • f: Find the probability that the teachers earn a total of over$400,000.
• g: Find the 90th percentile for an individual teacher’s salary.
• h: Find the 90th percentile for the average teachers’ salary.
• i: If we surveyed 70 teachers instead of 10, graphically, how would that change the distribution for X¯X¯ size 12{ {overline {X}} } {}?
• j: If each of the 70 teachers received a $3000 raise, graphically, how would that change the distribution for X¯X¯ size 12{ {overline {X}} } {}? ### Solution • a: X=X= size 12{X={}} {}the salary of a school teacher in a particular elementary school district • b: Xbar=Xbar= size 12{ ital "Xbar"={}} {}the average salary of 10 school teachers in a particular elementary school district • c: Xbar ~ N ( 44 , 000 , 6500 10 ) Xbar ~ N ( 44 , 000 , 6500 10 ) size 12{ ital "Xbar" "~" N $$"44","000", { {"6500"} over { sqrt {"10"} } }$$ } {} • d: SumX=SumX= size 12{ ital "SumX"={}} {}the total salaries of 10 school teachers in a particular elementary school district • e: SumX ~ N ( 10 44 , 000 , ( 10 6500 ) ) SumX ~ N ( 10 44 , 000 , ( 10 6500 ) ) size 12{ ital "SumX" "~" N $$"10" * "44","000", sqrt { \( "10" * "6500"$$ } \) } {} • f: P ( SumX > 400 , 000 ) = P ( Z > 1 . 9460 ) = 0 . 9742 P ( SumX > 400 , 000 ) = P ( Z > 1 . 9460 ) = 0 . 9742 size 12{P $$ital "SumX">"400","000"$$ =P $$Z> - 1 "." "9460"$$ =0 "." "9742"} {} • g: Find kk size 12{k} {} such that P(X<k)=0.90P(X<k)=0.90 size 12{P $$X<k$$ =0 "." "90"} {}; k=$52,330k=$52,330 size 12{k=$"52","330"} {}
• h: Find kk size 12{k} {} such that P(Xbar<k)=0.90P(Xbar<k)=0.90 size 12{P $$ital "Xbar"<k$$ =0 "." "90"} {}; k=$46,634k=$46,634 size 12{k=$"46","634"} {} • i: Since the standard deviation for XbarXbar size 12{ ital "Xbar"} {} is smaller if the sample size is 70, the distribution for XbarXbar size 12{ ital "Xbar"} {} is taller and thinner, graphically. • j: If each of the 70 teachers received a$3,000 raise, the distribution would have the same shape but would be shifted to the right by 3,000.

## Exercise 14

The distribution of income in some Third World countries is considered wedge shaped (many very poor people, very few middle income people, and few to many wealthy people). Suppose we pick a country with a wedge distribution. Let the average salary be $2000 per year with a standard deviation of$8000. We randomly survey 1000 residents of that country.

• a: In words, X=X= size 12{X={}} {}
• b: In words, X¯=X¯= size 12{ {overline {X}} ={}} {}
• c: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {}
• d: How is it possible for the standard deviation to be greater than the average?
• e: Why is it more likely that the average of the 1000 residents will be from $2000 to$2100 than from $2100 to$2200?

## Exercise 15

The average length of a maternity stay in a U.S. hospital is said to be 2.4 days with a standard deviation of 0.9 days. We randomly survey 80 women who recently bore children in a U.S. hospital.

• a: In words, X=X= size 12{X={}} {}
• b: In words, X¯=X¯= size 12{ {overline {X}} ={}} {}
• c: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {}
• d: In words, ΣX=ΣX= size 12{ΣX={}} {}
• e: ΣX ~ ΣX ~ size 12{ΣX "~" } {}
• f: Is it likely that an individual stayed more than 5 days in the hospital? Why or why not?
• g: Is it likely that the average stay for the 80 women was more than 5 days? Why or why not?
• h: Which is more likely:
• i: an individual stayed more than 5 days; or
• ii: the average stay of 80 women was more than 5 days?
• i: If we were to sum up the women’s stays, is it likely that, collectively they spent more than a year in the hospital? Why or why not?

### Solution

• a: X=X= size 12{X={}} {}the length of a maternity stay in a U.S. hospital, in days
• b: Xbar=Xbar= size 12{ ital "Xbar"={}} {} the average length of a maternity stay in a U.S. hospital, in days
• c: Xbar~N(2.4,0.980)Xbar~N(2.4,0.980) size 12{ ital "Xbar" "~" N $$2 "." 4, { {0 "." 9} over { sqrt {"80"} } }$$ } {} mean =2.4=2.4 size 12{ {}=2 "." 4} {} standard deviation =0.1006=0.1006 size 12{ {}=0 "." "1006"} {}
• d: SumX=SumX= size 12{ ital "SumX"={}} {} the total length of 80 maternity stays in U.S. hospitals, in days
• e: SumX~N(802.4,800.9)SumX~N(802.4,800.9) size 12{ ital "SumX" "~" N $$"80" * 2 "." 4, sqrt {"80"} * 0 "." 9$$ } {} mean =192=192 size 12{ {}="192"} {} standard deviation 8.04988.0498 size 12{8 "." "0498"} {}
• f: More than 5 days is more than at most 3 standard deviations above the mean so it is not very likely that an individual stays more than 5 days but it could happen.
• g: P(Xbar>5)=0P(Xbar>5)=0 size 12{P $$ital "Xbar">5$$ =0} {} so it is not likely at all that the average stay of women is more than 5 days.
• h: It is more likely for an individual to stay more than 5 days.
• i: Collectively, the sum, on the average, would be 802.4=192802.4=192 size 12{"80" * 2 "." 4="192"} {} days. Since 365 days is over twenty stdev.’s above the expected total of 192 days, it is not likely that, collectively, they spent more than a year in the hospital.

## Exercise 16

In 1940 the average size of a U.S. farm was 174 acres. Let’s say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940. (Source: U.S. Dept. of Agriculture)

• a: In words, X=X= size 12{X={}} {}
• b: In words, X¯=X¯= size 12{ {overline {X}} ={}} {}
• c: X ¯ ~ X ¯ ~ size 12{ {overline {X}} "~" } {}
• d: The IQR for X¯X¯ size 12{ {overline {X}} } {} is from __________ acres to __________ acres.

## Exercise 17

The stock closing prices of 35 U.S. semiconductor manufacturers are given below. (Source: Wall Street Journal)

8.625, 30.25, 27.625, 46.75, 32.875, 18.25, 5, 0.125, 2.9375, 6.875, 28.25, 24.25, 21, 1.5, 30.25, 71, 43.5,8.625, 30.25, 27.625, 46.75, 32.875, 18.25, 5, 0.125, 2.9375, 6.875, 28.25, 24.25, 21, 1.5, 30.25, 71, 43.5, size 12{8 "." "625, 30" "." "25, 27" "." "625, 46" "." "75, 32" "." "875, 18" "." "25, 5, 0" "." "125, 2" "." "9375, 6" "." "875, 28" "." "25, 24" "." "25, 21, 1" "." "5, 30" "." "25, 71, 43" "." "5,"} {} 49.25, 2.5625, 31, 16.5, 9.5, 18.5, 18, 9, 10.5, 16.625, 1.25, 18, 12.875, 7, 12.875, 2.875, 60.25, 29.25 49.25, 2.5625, 31, 16.5, 9.5, 18.5, 18, 9, 10.5, 16.625, 1.25, 18, 12.875, 7, 12.875, 2.875, 60.25, 29.25 size 12{" 49" "." "25, 2" "." "5625, 31, 16" "." "5, 9" "." "5, 18" "." "5, 18, 9, 10" "." "5, 16" "." "625, 1" "." "25, 18, 12" "." "875, 7, 12" "." "875, 2" "." "875, 60" "." "25, 29" "." "25"} {}

• a: In words, X=X= size 12{X={}} {}
• b: x¯=x¯= size 12{ {overline {x}} ={}} {}__________ sx=sx= size 12{s rSub { size 8{x} } ={}} {} __________ n=n= size 12{n={}} {}__________
• c: Construct a histogram of the distribution of the averages. Start at x=0.0005x=0.0005 size 12{x= - 0 "." "0005"} {}. Make bar widths of 10.
• d: In words, describe the distribution of stock prices.
• e: Randomly average 5 stock prices together. (Use a random number generator.) Continue averaging 5 pieces together until you have 10 averages. List those 10 averages.
• f: Use the 10 averages from (e) to calculate: x¯=x¯= size 12{ {overline {x}} ={}} {} __________ sx¯=sx¯= size 12{ {overline {s rSub { size 8{x} } }} ={}} {}__________
• g: Construct a histogram of the distribution of the averages. Start at x=0.0005x=0.0005 size 12{x= - 0 "." "0005"} {}. Make bar widths of 10.
• h: Does this histogram look like the graph in (c)?
• i: In 1 - 2 complete sentences, explain why the graphs either look the same or look different?
• j: Based upon the theory of the Central Limit Theorem, X¯~X¯~ size 12{ {overline {X}} "~" } {}_______________

### Solution

• a: X=X= size 12{X={}} {} a stock closing price for the business week ending Jan. 27, 1995
• b: Sample mean =$20.71=$20.71 size 12{ {}=$"20" "." "71"} {}; sample standard deviation =$17.31=$17.31 size 12{ {}=$"17" "." "31"} {}; sample size =35=35 size 12{ {}="35"} {}
• c:
• d: Exponential distribution
• e: (will vary)
• f: (will vary)
• g: (will vary)
• h: No.
• i: The graphs look different because the second one is a graph of averages and will tend to a bell shape.
• j: Xbar~N(20.71,17.315)Xbar~N(20.71,17.315) size 12{ ital "Xbar" "~" N $$"20" "." "71", { {"17" "." "31"} over { sqrt {5} } }$$ } {} by the Central Limit Theorem

## Exercise 18

• a: In words, X=X= size 12{X={}} {}
• b: μx=μx= size 12{μ rSub { size 8{x} } ={}} {} __________ σx=σx= size 12{σ rSub { size 8{x} } ={}} {} __________ n=n= size 12{n={}} {}__________
• c: Construct a histogram of the distribution. Start at x=0.50x=0.50 size 12{x= - 0 "." "50"} {}. Make bar widths of $5. • d: In words, describe the distribution of stock prices. • e: Randomly average 5 stock prices together. (Use a random number generator.) Continue averaging 5 pieces together until you have 15 averages. List those 15 averages. • f: Use the 15 averages from (e) to calculate: x¯=x¯= size 12{ {overline {x}} ={}} {}__________ sx¯=sx¯= size 12{ {overline {s rSub { size 8{x} } }} ={}} {}__________ • g: Construct a histogram of the distribution of the averages. Start at x=0.50x=0.50 size 12{x= - 0 "." "50"} {}. Make bar widths of$5.
• h: Does this histogram look like the graph in (c)? Explain any differences.
• i: In 1 - 2 complete sentences, explain why the graphs either look the same or look different?
• j: Based upon the theory of the Central Limit Theorem, X¯~X¯~ size 12{ {overline {X}} "~" } {}_______________

Try these multiple choice questions.

Questions 19 and 20 refer to the following: The time to wait for a particular rural bus is distributed uniformly from 0 to 75 minutes. 100 riders are randomly sampled to learn how long they waited.

## Exercise 19

The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is:

• A: 315.0
• B: 40.3
• C: 38.5
• D: 65.2

B

## Exercise 20

Would you be surprised, based upon numerical calculations, if the sample average wait time (in minutes) for 100 riders was less than 30 minutes?

• A: Yes
• B: No
• C: There is not enough information.

A

## Exercise 21

Which of the following is NOT TRUE about the distribution for averages?

• A: the mean, median and mode are equal
• B: the area under the curve is one
• C: the curve never touches the x-axis
• D: the curve is skewed to the right

### Solution

D

For questions 22 – 23, use the following information: The average cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $2.59 and a standard deviation of$0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.

## Exercise 22

The distribution to use for the average cost of gasoline for the 16 gas stations is

• A: X ¯ ~ N ( 2 . 59 , 0 . 10 ) X ¯ ~ N ( 2 . 59 , 0 . 10 ) size 12{ {overline {X}} "~" N $$2 "." "59",0 "." "10"$$ } {}
• B: X ¯ ~ N ( 2 . 59 , 0 . 10 16 ) X ¯ ~ N ( 2 . 59 , 0 . 10 16 ) size 12{ {overline {X}} "~" N $$2 "." "59", { {0 "." "10"} over { sqrt {"16"} } }$$ } {}
• C: X ¯ ~ N ( 2 . 59 , 0 . 10 16 ) X ¯ ~ N ( 2 . 59 , 0 . 10 16 ) size 12{ {overline {X}} "~" N $$2 "." "59", { {0 "." "10"} over {"16"} }$$ } {}
• D: X ¯ ~ N ( 2 . 59 , 16 0 . 10 ) X ¯ ~ N ( 2 . 59 , 16 0 . 10 ) size 12{ {overline {X}} "~" N $$2 "." "59", { {"16"} over {0 "." "10"} }$$ } {}

B

## Exercise 23

What is the probability that the average price for 16 gas stations is over \$2.69?

• A: almost zero
• B: 0.1587
• C: 0.0943
• D: unknown

A

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