Question

Math 125: May 18, 2018 To get full credit you must show your work. No work, no credit. Each question is worth about 9 points. 1. The figure below is a histogram showing the distribution of blood pressure for about 14,000 women in a drug study. Use the histogram to answer the following questions: (a) Is the percentage of women with blood pressures between 90 mm and 160 mm around 1%, 50%, or 99%? (b) In which interval are there more women: 135-140 mm or 140-150 mm? (c) Which interval is more crowded: 135-140 mm or 140-150 mm? (d) On the interval 125-130 mm, the height of the histogram is about 2.1% per rnm. what percentage of the women had blood pressures in this class interval? %per min 0 90 100 110 120 130 140 150 160 BLOOD PRESSURE (MM) 2. Among freshmen at a certain university, scores on the Math SAT followed the normal curve, with an average of 550 and an SD of 100. Fill in the blanks; explain briefly (a) A student who scored 400 on the Math SAT was at about theth percentile of the score distribution. (b) To be at the 75th percentile of the distribution, a student needed a score of about points on the Math SAT.

Answer 1

1.

(a) The histogram is plotted per mm. So, for example at 90-91 mm, there is around 1/2 % of women. Also, for 91-92 mm, 92-93 mm, upto 99-100 mm. At 100-101 mm, it is around 2%. Summing up the percentages per mm gives us nearly around 99% of the women.

(b) There are more women in 140-150 mm interval than in 135-140 mm interval. This is because the area under the 140-150 mm is more.

(c) The 135-140 mm interval is more crowded. This is because the height of this interval is more.

(d) The 125-130 mm interval contains 5 mm. So, this class represents 2.1 %/mm * 5 mm = 10.5% of women.

2.

(a) We first compute the z-score for the score of 400.

$\\ \mu = 550\\ \sigma = 100\\ \texttt{z-score}=\frac{x-\mu}{\sigma}=\frac{400-550}{100}=\mathbf{-1.5}$

Using a z-table, we get the corresponding fraction as 0.0668 = 6.68% = 7%.

So, a score of 400 is about 7th precentile of the score distribution.

(b) For 75th percentile of the distribution, we look up the z-table for fraction of 75% = 0.75. The corresponding z-score is 0.67. So, we have:

$\\ \texttt{z-score}=\frac{x-\mu}{\sigma}\\ \therefore x = \mu + \texttt{z-score}\cdot \sigma=550 + 0.67\cdot 100=\mathbf{617}$