MATH 1040 Practice Midterm 3 Solutions: Probability Distributions and Confidence Intervals, Exams of Mathematics

Download MATH 1040 Practice Midterm 3 Solutions: Probability Distributions and Confidence Intervals and more Exams Mathematics in PDF only on Docsity! MATH 1040 PRACTICE MIDTERM 3 SOLUTIONS 1. PROBLEM 1 A population follows a uniform probability model. If the probability density function is given by f(x) = } for 1 < x < 6 compute P(X < 3), P(X <3), and P(X > 2). Interpret your results in terms of area. Solution 1. We compute: (1) P(X >2)=(6-2)- cur cue Figure 1 illustrates the geometric interpretation. The probability we have just computed is the area of the shaded region A. 2. PROBLEM 2 Suppose that X is a normal random variable with mean p= 3 and standard deviation o = 2. Compute P(X < 5), P(X > 3), and P(X <2). Solution 2. We make use of the standard normal distribution. Using table IV we compute: 1/5 FIGuRE 1. Illustration for problem 1 1 2 MATH 1040 PRACTICE MIDTERM 3 SOLUTIONS P (X ≤ 5) = P ( Z ≤ 5− 3 2 ) = P (Z ≤ 1) ≈ 0.8413 P (X ≥ 3) = 1− P (X < 3) = 1− P ( Z < 3− 3 2 ) = 1− 1 2 = 1 2 P (X < 2) = P ( Z < 2− 3 2 ) = P ( Z < −1 2 ) ≈ 0.3085 (2) We say approximately equal in the first and third calculations since the table is inexact (as is a calculator). The approximations are to four decimal places. In the second calculation, we simply used the symmetry properties of the normal distribution. In this situation, the calculation is exact. 3. Problem 3 Compute zα when α = 0.9, 0.2, and 0.1. Solution 3. Again, we use table IV: P (Z ≥ z0.9) = 0.9 P (Z < z0.9) = 1− 0.9 = 0.1 z0.9 ≈ −1.28. (3) By symmetry, z0.1 = −z0.9 ≈ 1.28. Also by table IV: P (Z ≥ z0.2) = 0.2 P (Z < z0.2) = 1− 0.2 = 0.8 z0.2 ≈ 0.84. (4) The approximations are to two decimal places. 4. Problem 4 Suppose that X is a normal random variable with mean µ = 0 and standard deviation σ = 6. Compute a such that P (−a ≤ X ≤ a) = 1 2 . MATH 1040 PRACTICE MIDTERM 3 SOLUTIONS 5 9. Problem 9 A simple random sample of size n = 10 is drawn from a population known to be normally distributed. The sample standard deviation is s = 5 and the sample mean is x = 10. Construct a 98% confidence interval about x using the t-table. Solution 9. We have 9 degrees of freedom and α0.02. From table V, we see that the confidence interval is given by x̄± t0.01 · s√ n ≈ 10± 2.821 · 5√ 10 ≈ 10± 4.460 (10) The second approximation is an approximation of the first approxima- tion and in that context is correct to 4 decimal places. Notice how much larger our error is then in the comparable situation for problem 9. Knowing the population standard deviation in advance can make a big difference. 10. Problem 10 The sample size required to obtain a (1−α)100% confidence interval for p with a margin of error E is given by (11) n = p̂(1− p̂) (zα 2 E )2 where p̂ is a prior estimate of p. We are given that p̂ = 5/6. What does the sample size have to be if α = 0.05 and the margin of error is 0.1? Solution 10. We compute: (12) p̂(1− p̂) (zα 2 E )2 = 5 6 · 1 6 · (z0.025 0.1 )2 ≈ 5 36 · (1.96 0.1 )2 ≈ 53.36. Hence, we need a sample size of at least 54. 11. Problem 11 A simple random sample of size n = 10 is drawn from a population known to be normally distributed. The sample standard deviation is s = 5. Construct a 98% confidence interval about σ2. The inequality (13) (n− 1)s2 χ2α 2 < σ2 < (n− 1)s2 χ21−α 2 may be of use here. 6 MATH 1040 PRACTICE MIDTERM 3 SOLUTIONS Solution 11. Notice that α0.02 and that we have 9 degrees of freedom. From table VI, our two χ2 values are 21.666 and 2.088. The lower bound of our confidence interval is (14) 9 · 52 21.666 ≈ 10.3849 while the upper bound is (15) 9 · 52 2.088 ≈ 107.7586 with both approximations being taken to four decimal places. 12. Problem 12 A random variable is normally distributed with mean 2 and standard deviation 1. Find the area under the curve (that is the graph of the probability density function for the random variable) between x = −2 and x = 2. Solution 12. Let A be the area in the question. Noting the relationship between probability and area we compute: (16) A = P (−2 < X < 2) = P (X < 2)−P (X < −2) = 1 2 −P (Z < −4) ≈ 1 2 .