Download 3 Problems on the Introductory Biostatistics - Exam | PUBHLTH 540 and more Exams Biostatistics in PDF only on Docsity! PubHlth 540 Introductory Biostatistics Page 1 of 3 Topic 5 – The Normal Distribution Week #8 - Practice Problems SOLUTIONS 1. Suppose the distribution of GRE scores satisfies the assumptions of normality with a mean score of μ=600 and a standard deviation of σ=80. a. What is the probability of a score less than 450 or greater than 750? Answer: .0608 Solution: Probability { score < 450 OR score > 750 } [ ] [ ] [ ] [ ] [ ] ( ) = pr X < 450 + pr X > 750 450-600 750-600= pr Z < + pr Z > 80 80 = pr Z < -1.875 + pr Z > +1.875 = 2pr Z > +1.875 = 2 .0304 = .0608 ⎡ ⎤ ⎡⎛ ⎞ ⎛ ⎜ ⎟ ⎜ ⎤⎞ ⎟⎢ ⎥ ⎢⎝ ⎠ ⎝ ⎥⎠⎣ ⎦ ⎣ ⎦ b. What proportion of students have scores between 450 and 750? Answer: .9392 Solution: Proportion of students with scores between 450 and 750 [ ]= pr 450 < X < 750 =1 - pr[ x < 450 or X > 750] = −1 0608. =.9392 c. What score is equal to the 95th percentile? Answer: 731.2 Solution: For Z∼Normal(0,1) [ ].95pr Z < 1.645 = .95 From X - μZ = σ substitute .95 X - 6001.645 = 80 ….docu\Wk8_solutions.doc PubHlth 540 Introductory Biostatistics Page 2 of 3 Thus, X Z. .95 95= +σ μ = +80 1645 600a f . 731.6= 2. The Chapin Social Insight Test evaluates how accurately the subject appraises other people. In the reference population used to develop the test, scores is normally distributed with mean μ=25 and standard deviation σ=5. The range of possible scores is 0 to 41. a. What proportion of the population has scores below 20 on the Chapin test? Answer: .1587 Solution: 20 - 25pr(X < 20) = pr Z < = pr[Z < -1]=.1587 5 ⎡ ⎤ ⎢ ⎥⎣ ⎦ b. What proportion has scores below 10? Answer: .0014 Solution: 10 - 25pr(X < 10) = pr[ Z < ] = pr[ Z < -3 ] = .0014 5 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: ( 0.6745) .25pr Z > = Thus, X = σZ + μ ( )[ ]5 0.6745 25= + 28.3725= 3. 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: First obtain an interval for Z∼Normal(0,1) [ ]pr -1.96 < Z < +1.96 = .95 Next, recall that the standard error, SE, of X , is related to σ via σse[X] = n And the mean of X is E [X] = μ Thus, the standardization formula can be manipulated to yield a formula for X in terms of Z. ….docu\Wk8_solutions.doc
