Download Statistics for Engineering - Quiz - Homework | STAT 4706 and more Quizzes Statistics in PDF only on Docsity! STAT-4706 Homework # 1 Fall 2012 You must show all of your work for each problem in order to receive full credit. Minitab should be used when specified. HW Due 09/13/2012. 1. (25)The lengths of time, in minutes, that 10 patients waited in a doctor’s office before receiving treatment were recorded as follows: 5, 11, 9, 5, 10, 15, 6, 10, 5, and 10. Treating the data as a random sample, find the: a. (5)Sample mean and sample variance. x́= 1 n ∑ i=1 n x i= 5+11+9+…+10 10 =8.6 s2= 1 n−1 ∑ i=1 n (x i− x́ ) 2 = (5−8.6 )2+(11−8.6 )2+(9−8.6 )2+…+(10−8.6 )2 9 = 12.96+5.76+0.16+…+1.96 9 =10.933 Variance = S2 = 10.933 ** SD = √S2=√10.93=3.31 (not part of homework) b. (5)Median. i. First order the data : 5 5 5 6 9 10 10 10 11 15 ii. For a data set of 10 observations (n even), the median (m) is calculated as m= x(5)+x (6 ) 2 = 9+10 2 The median is m = 9.5 c. (3)Mode. Two modes; 5 and 10 minutes. d. (6)Upper and lower quartiles and IQR. 5 5 5 6 9 10 10 10 11 15 Min Q1 Q2= Median Q3 Max Q0 = Min = 5 Q1 = 5 Q2= Median =9.5 Q3= 10 Q4 = Max =15 IQR = Q3 - Q1 = 5 e. (6)Manually, construct a boxplot of the data. Q1 Q2 Q3 5 7 9 11 13 15 17 1 STAT-4706 Homework # 1 Fall 2012 2. (30)The heights of 1000 students are approximately normally distributed with a mean of 174.5 cm. and SD of 6.9 cm. Suppose 200 random samples of size 25 are drawn from this population and the means recorded to the nearest tenth of a centimeter. Determine a. (4)The mean and SD of the sampling distribution of X́ . μ X́ =μ=174.5 cm. σ X́= σ √n = 6.9 √25 =1.38 b. (14)The number of sample means that fall between 172.5 and 175.8 centimeters inclusive. 0.30 0.25 0.20 0.15 0.10 0.05 0.00 X D e n si ty 172.45 0.7673 175.85174.5 Distribution Plot Normal, Mean=174.5, StDev=1.38 Z1= (172.45−174.5) 1.38 =−1.49 Z2= (175.85−174.5) 1.38 =0.98 so, P(172.45 < X́ < 175.85) = P(-1.49 < Z < 0.98) = 0.8365 – 0.068 = 0.7684 Therefore, the number of sample means between 172.5 and 175.8 inclusive is (200)*(0.7684) = 154. c. (12)The number of sample means falling below 172.0 cm. 0.30 0.25 0.20 0.15 0.10 0.05 0.00 X D e n si ty 171.95 0.03231 174.5 Distribution Plot Normal, Mean=174.5, StDev=1.38 Z1= (171.95−174.5) 1.38 =−1.85 So, P(X́ < 171.95) = P(Z < -1.85) = 0.0322 Therefore, about (200)(0.0322) = 6 sample means fall below 172.0 cm 2
