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BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 1 of 16 Unit 7 – The Normal Distribution Homework SOLUTIONS 1. This exercise gives you practice in calculating probabilities under the standard normal curve. See lecture notes for unit 7 page 15. A good url to use is http://davidmlane.com/hyperstat/z_table.html Recall the convention of using the letter Z to represent a random variable that is distributed standard normal. Find the proportion of observations from a standard normal distribution that satisfies each of the following statements. a. b. c. d. e. f. g. h. Z < 2.85 Z > 2.85 Z > -1.66 -1.66 < Z < 2.85 Z < -2.25 Z > -2.25 Z > 1.77 -2.25 < Z < 1.77 Notes – I used http://davidmlane.com/hyperstat/z_table.html #1a. Pr (Z < 2.85) = .9978 sol_normal.docx BIOSTATS 540 – Fall 2016 #1b. Pr (Z > 2.85) = .0022 #1c. Pr (Z > -1.66) = .9515 sol_normal.docx Introductory Biostatistics Page 2 of 16 BIOSTATS 540 – Fall 2016 #1d. Pr (-1.66 < Z < 2.85) = .9494 #1e. Pr (Z < -2.25) = .0122 sol_normal.docx Introductory Biostatistics Page 3 of 16 BIOSTATS 540 – Fall 2016 #1f. Pr (Z > -2.25) = .9878 #1g. Pr (Z > 1.77) = .0384 sol_normal.docx Introductory Biostatistics Page 4 of 16 BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 5 of 16 #1h. Pr (-2.25 < Z < 1.77) = .9494 2. This exercise gives you practice in calculating probabilities under normal curves with nonzero mean and non-unit variance. The same url will work for this exercise too. http://davidmlane.com/hyperstat/z_table.html The height, X, of young American women is distributed normal with mean µ=65.5 and standard deviation σ=2.5 inches. Find the probability of each of the following events. a. X < 67 b. 64 < X < 67 #2a. .7257 ⎡⎛ X-µ ⎞ ⎛ 67-µ ⎞ ⎤ pr(X < 67) = pr ⎢⎜ ⎟<⎜ ⎟⎥ σ ⎠ ⎝ σ ⎠⎦ ⎣⎝ ⎡ ⎛ 67-65.5 ⎞ ⎤ ⎟⎥ ⎝ 2.5 ⎠ ⎦ . = pr ⎢ Z < ⎜ ⎣ = pr [ Z < .6] = .7257 sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics #2b. .4515 ⎡⎛ 64-65.5 ⎞ ⎛ 67-65.5 ⎞ ⎤ pr ( 64 < X < 67 ) = pr ⎢⎜ ⎟ <Z< ⎜ ⎟⎥ ⎝ 2.5 ⎠ ⎦ ⎣⎝ 2.5 ⎠ . = pr [-0.6 < Z < +0.6] =.4515 sol_normal.docx Page 6 of 16 BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 7 of 16 3. This exercise gives you additional practice in calculating probabilities under normal curves with non-zero mean and non-unit variance. Suppose that, in a certain population, the distribution of GRE scores is normal with mean µ=600 and standard deviation σ=80. a. What is the probability of a score less than 450 or greater than 750? Answer: .0608 Solution: Define the random variable X = GRE score. Thus, X is distributed normal with mean µ=600 and standard deviation σ=80. We write this more compactly as X ~ Normal (µ=600, σ=80). à Probability { score < 450 OR score > 750 } = pr [ X < 450] + pr [ X > 750] = .0608 Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1) Click on button for “area from value” (2) OUTSIDE (3) values 450 and 750 (4) mean = 600 and (5) SD=80. sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 8 of 16 b. What proportion of students has scores between 450 and 750? Answer: .9392 Solution: “Proportion” of students with scores between 450 and 750 à we want: = pr [ 450 < X < 750] =.9392 Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1) Click on button for “area from value” (2) BETWEEN (3) values 450 and 750 (4) mean = 600 and (5) SD=80. sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 9 of 16 c. What score is equal to the 95th percentile? Answer: 731.2 There are at least two solutions to this question: Solution I – Simple “plug in” variety Solution II – 2 step solution that re-enforces the concepts.. Step 1: Obtain the 95th percentile for Z∼Normal(0,1). Call this Z.95 Step 2: Use Z.95 and the formula on page 19 of the course notes to obtain X.95 Solution I: Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1) Click on button for ‘value from an area” (2) area = .95 (3) mean = 600 and (4) SD=80. Don’t forget to click RECALCULATE. sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 10 of 16 Solution II Step 1: Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1) Click on button for “value from an area” (2) BELOW, (3) area = .95 (4) mean = 0 and (5) SD=1. Solution II Step 2: Use the formula on page 19 of the unit 5 notes with the following inputs: (1) Z.95 = 1.645 (2) µ = 600 and σ = 80 X .95 = σZ.95 + µ = ( 80 )[1.645 ] + 600 = 731.6 sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 11 of 16 4. The Chapin Social Insight Test evaluates how accurately the subject appraises other people. In the reference population used to develop the test, Chapin Social Insight Test scores are distributed normal with mean µ=25 and standard deviation σ=5. a. What proportion of the population has scores below 20 on the Chapin test? Answer: .1587 Solution: The solution for the “proportion of the population” is a probability calculation. Define the random variable X = Chapin Social Insight Test Score. X is distributed Normal (µ=25, σ=5). Want: pr(X < 20) = .1587 Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1) Click on button for “area from a value” (2) BELOW (3) value =20 (4) mean = 25 and (4) SD=5. sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 12 of 16 b. What proportion has scores below 10? Answer: .0014 Solution: This is similar to “a”. à The solution for the “proportion of the population” is a probability calculation. X is distributed Normal (µ=25, σ=5). Want: pr(X < 10) = .0014 url used: http://davidmlane.com/hyperstat/z_table.html sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 13 of 16 c. How high a score must you have in order to be in the top quarter of the population in social insight? Answer: 28.35 Solution: Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1) Click on button for ‘value from an area” (2) ABOVE (3) area = .25 (4) mean = 25 and (5) SD=5. sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 14 of 16 5. A normal distribution has mean µ=100 and standard deviation σ=15 (for example, IQ). Give limits, symmetric about the mean, within which 95% of the population would lie: Solution: This exercise is asking you to work with the following characteristic of the Normal distribution: If X1, X2, …. Xn are a simple random sample, each distributed Normal(µ, σ2) Then the sample mean of n observations is distributed Normal((µ, σ2/n) Tip! The “David Lane” calculator does not have a special box labeled SE So, to use the “David Lane” calculator for the normal distribution of a sample mean, it is necessary to input the value of σ 2 / n in the box “SD” a) Individual observations Answer: 70.6, 129.4 “Individual observations” à want to set the “David Lane” boxes as follows: “MEAN” = µ=100 “SD” = σ = 15. sol_normal.docx BIOSTATS 540 – Fall 2016 Introductory Biostatistics b) Means of 4 observations Answer: 85.3, 114.7 “Means of 4 observations” à want to set the “David Lane” boxes as follows: “MEAN” = µ=100 “SD” = SE = √(σ 2/n) = σ/√4 = 15/2 = 7.5 c) Means of 16 observations Answer: 92.65, 107.35 “Means of 16 observations” à want to set the “David Lane” boxes as follows: “MEAN” = µ=100 “SD” = SE = √(σ 2/n) = σ/√16 = 15/4 = 3.75 sol_normal.docx Page 15 of 16 BIOSTATS 540 – Fall 2016 Introductory Biostatistics Page 16 of 16 d) Means of 100 observations Answer: 97.06, 102.94 “Means of 100 observations” à want to set the “David Lane” boxes as follows: “MEAN” = µ=100 “SD” = SE = √(σ 2/n) = σ/√100 = 15/10 = 1.5 sol_normal.docx