Question

1153 – Chapter 12 & 5 HW Questions 1. ACT scores follow an approximately normal distribution. One year, the mean score was 18.2 with standard deviation 5.8. Round proportions to 4 decimal places and scores to the appropriate integer. (a) What proportion of students scored below 20? Above 25? Between 20 and 30? (b) What score did someone at the 80th percentile have? (c) Describe the top 10% of ACT scores: (d) Describe the middle 50% of ACT scores:

2. At a zoo, Biteyfloofers have a mean length of 11” and standard devation 2.5”, while Fluffersnappers have a mean length of 10” and standard deviation 2”. Both follow an approx. normal distribution. (a) Which is more unusual, a Fluffersnapper that is 12” long or a Biteyfloofer that is 12”? (b) Which would seem longer relative to their populations, a 9” Fluffersnapper or a 9.5” Biteyfloofer? (c) Use the empirical rule to compare the middle 95% of heights for each creature.

3. For each situation, decide if it would be reasonable to use the normal distribution to describe the sampling distribution. Supply a brief explanation for each. (a) A polling company takes a SRS of 1000 voters and calculates the proportion who plan to vote in favor of Proposition 60. Previous polls suggest 26% are in favor of Prop. 60. (b) Brian rolls a fair 6-sided die 50 times and records the number of times “2” comes up.

4. A 2011 Gallup poll found that 76% of Americans believe that high achieving high school students should be recruited to become teachers. This poll was based on a random sample of 1002 Americans. (a) Find a 90% confidence interval for the proportion of Americans who would agree with this. (b) Interpret your interval in this context. (c) Explain what “90% confidence” means in this context. (d) Do these data refute a pundit’s claim that at least 2/3 of Americans believe this statement. Explain.

5. A 95% confidence interval for a politician’s level of support is given by (0.29, 0.46). Which is correct? Why? (a) There’s a 95% chance that the interval correctly captured the politician’s level of support for those sampled, i.e. that between 29% and 46% of the sample support the politician. (b) 95% of all samples will produce values between 29% and 46%. (c) There’s a 95% chance that the interval correctly captured the politician’s level of support, i.e. that between 29% and 46% of the constituents support the politician. (d) 95% of the time, between 29% and 46% of constituents will support the politician.

6. For each scenario, indicate if the chapter 13 confidence interval is appropriate, and briefly explain why. (a) A prison has records on all 385 of its inmates, and finds that 59% have been vaccinated against pararibulitis. (b) A random sample of 1178 inmates in prisons across Oklahoma finds that 59% have been vaccinated against pararibulitis. (Note: Oklahoma has more than 62,000 inmates) (c) A prison doctor notes that, of the last 86 inmates to receive medical care, 59% have been vaccinated against pararibulitis.

Answer 1

SOLUTION 1a: The mean score was 18.2 with standard deviation = 5.8.

i) Proportion of students scored below 20:

P ( X<20 )=P ( X−μ<20−18.2 )=P ((X−μ)/σ<(20−18.2)/5.8)

Since (x−μ)/σ=Z and (20−18.2)/5.8=0.31 we have:

P (X<20)=P (Z<0.31)

Use the standard normal table to conclude that:

P (Z<0.31)=0.6217

ii) Proportion of students scored Above 25.

P ( X>25 )=P ( X−μ>25−18.2 )=P ((X−μ)/σ>(25−18.2)/5.8)

Since Z=(x−μ)/σ and (25−18.2)/5.8=1.17 we have:

P ( X>25 )=P ( Z>1.17 )

Use the standard normal table to conclude that:

P (Z>1.17)=0.121

iii) Proportion of students scored Between 20 and 30:

P ( 20<X<30 )=P ( 20−18.2< X−μ<30−18.2 )=P ((20−18.2)/5.8<(X−μ)/σ<(30−18.2)/5.8)

Since Z=(x−μ)/σ , (20−18.2)/5.8=0.31 and (30−18.2)/5.8=2.03 we have:

P ( 20<X<30 )=P ( 0.31<Z<2.03 )

Use the standard normal table to conclude that:

P ( 0.31<Z<2.03 )=0.3571

b) Z score corresponding to 80th percentile score is 0.842

Z= X-mean/sigma

Z*sigma= X-mean

0.842*5.8=X-18.2

X= 23.0836

c) The Top 10% means Zscore corresponding to 0.9 is 1.28

Z= X-mean/sigma

Z*sigma= X-mean

1.28*5.8=X-18.2

X= 25.624

d) the middle 50% of ACT scores

We wish to find the tick-marks which separate 25% and 75% of the area (0.2500 and 0.75000) to the left.

The closest entries are z= −0.68 and 0.68

? = ?? + ? = 5.8(±0.68) + 18.2 = 14.256 and 22.144 ACT SCORES.

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