Math 3339 Homework 5 (Sections 5.5 & 6.5)

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1 Math 3339 Homework 5 (Sections 5.5 & 6.5) Name: PeopleSoft ID: Instructions: Homework will NOT be accepted through or in person. Homework must be submitted through CourseWare BEFORE the deadline. Print out this file and complete the problems. Use blue or black ink or a dark pencil. Write your solutions in the space provided. You must show all work for full credit. Submit this assignment at under Assignments" and choose WH5. Total possible points: Section 5.5, Problem 1 Let Z~Norm(0, 1). Use the normal table and also R s pnorm function to find (a) P(Z 1.45) (b) P(Z > -1.28) (c) P( Z < 1.036) (d) P(Z > 0.836) a) Table: , R: b) Table: = , R: c) Table: = , R: d) Table: = , R: Section 5.5, problem 2 (a) P(X 6.13), X ~ Norm(1, 4) Table: P(Z ) = R: pnorm(6.13,1,4) = (b) P r(x > -2.35), X ~ Norm(-1, 2) Table: P(Z > ) = R: 1 pnorm(-2.35,-1,2) = (c) Pr( < X 7.682), X ~ Norm(2.5, 5) Table: P( < ) = = R: pnorm(7.682,2.5,5) pnorm(-.872,2.5,5) = (d) P r(x > 0.698), X ~ Norm(-2, 4) Table: P(Z > ) = = R: 1 pnorm(0.698,-2,4) =

2 3. Section 5.5, Problem 3 (a) The 90th percentile of Norm(0, 5). Table: Z.9 = 1.28, X = 1.28(5) + 0 = 6.4 R: qnorm(.9,0,5) = (b) The 15th percentile of Norm(1, 3). Table: Z.15 = -1.04, X = -1.04(3) + 1 = R: qnorm(0.15,1,3) = (c) The interquartile range, i.e., the distance from the first to third quartiles of Norm(μ, σ). Table: Z.25 = -0.67, Z0.75 = 0.67, Q1 = -.67σ + μ, Q3 = 0.67σ + μ IQR = 0.67σ +μ +0.67σ - μ = 1.34σ R: Z0.25 = , Z0.75 = IQR = σ 4. *Human body temperatures for healthy individuals have approximately a Normal distribution with mean F and standard deviation 0.75 F. a. Find the 90 th percentile for temperatures of healthy individuals. b. Find the 5 th percentile for temperatures of healthy individuals. c. Determine the first quartile. a. qnorm(0.9, 98.25,.75) = b. qnorm(0.05,98.25,.75) = c. Q1 = qnorm(0.25,98.25,.75) =

3 5. *If adult female heights are Normally distributed, what is the probability that the height of a randomly selected woman is a. Within 1.5 SDs of its mean value? b. Farther than 2.5 SDs from its mean value? c. Between 1 and 2 SDs from its mean values? Number of standard deviations from the mean is the definition of the z-sore. a. P(-1.5 < Z < 1.5) = pnorm(1.5) pnorm(-1.5) = b. P(Z < or Z > 2.5) = 2 * pnorm(-2.5) = c. P(1 < Z < 2) = 2*(pnorm(2) pnorm(1)) = 2* =

4 6. *The inside diameter of a randomly selected piston ring is a random variable with mean value 12 cm and standard deviation 0.04 cm. a. If XX is the sample mean diameter for a random sample of n = 16 rings, where is the sampling distribution of XX centered, and what is the standard deviation of the XX distribution? b. Answer the questions posted in part (a) for a sample size of n =64 rings. c. For which of the two random samples, the one of part (a) or the one of part (b), is XX more likely to be within 0.01 cm of 12 cm? Explain your reasoning. a. Center of XX is µ = 12, SSSS(XX ) = SSSS(XX) = 0.04 = 0.01 nn 4 b. Center of XX is µ = 12, SSSS(XX ) = SSSS(XX) = 0.04 = nn 8 c. In part b because that is two standard deviation from the mean whereas for part a 0.01 would be one standard deviations from the mean. That is by the empirical rule: a) Approximately 68% of the means would be within 0.01 of the mean 12. b) Approximately 95% of the means would be within 0.01 of the mean 12.

5 7. *Refer to the previous problem (6). Suppose the distribution of the diameter is normal. a. Calculate PP(11.99 XX 12.01) when n = 16. b. How likely is it that the sample mean diameter exceeds when n = 25? a. P(11.99 XX 12.01) = pnorm(12.01,12,0.01) pnorm(11.99,12,0.01) = b. n = 25, SD(XX ) =.04 = 0.008; P(X > 12.01) = 1 pnorm(12.01,12,.008) = or 10.57% 25

6 8. * Suppose only 70% of all drivers in a certain state regularly wear a seat belt. A random sample of 500 drivers is selected. Using the Normal approximation, what is the probability that a. Between 320 and 370 (inclusive) of the drivers in the sample regularly wear a seat belt? b. Fewer than 325 of those in the sample wear a seatbelt. c. Answer parts a and b using pbinomial in R. Do you get the same answers? n = 500; p = 0.7 so µ = 500(.7) = 350 and σσ = 500(. 7)(.3) = a. P(320 X 370) = P(X 370) P(X 319) = pnorm(370.5,350,10.247) pnorm(319.5,350,10.247) = b. P(X < 325) = P(X 324) = pnorm(324.5,350,10.247) = c. 1) pbinom(370,500,.7) pbinom(319,500,.7) = ) pbinom(324,500,.7) = Close answers within thousdandths.