HP340: Statistical Methods, Spring 2015Due Date: 3:30PM, Tuesday Feb 10, 2015 Homework 3Total Points: 40Instructions: Clearly circle the letter of the best answer for multiple-choice questions and show your work when calculations are required. Late assignments will not be accepted.1. (8 points) a) (2 points) The standard normal distribution has a variance equal to ____ and a median equal to ____.A. 0; 1B. the interquartile range; 0C. 0 ; the z scoreD. 1; 0 E. 0 ; the raw scoreb) (2 points) When a variable is normally distributed, _____ of the values are expected to lie within two standard deviations of the median.A. 68.26%B. 95.44%C. 99.74%D. 100%E. Cannot be determined without knowing the mean.c) (4 points, 1 point each) True or False. Given a normal distribution with μ=1 and σ=1: A. The frequency of scores greater than 0 is the same as the frequency of scores less than 0. FALSEB. The percentage of scores greater than 1 is smaller than the percentage of scores lessthan 1. FALSEC. Approximately 95% of the scores fall between 0 and 2. FALSED. This is a standard normal distribution. FALSE2. (10 points) A library wants to determine the effectiveness of their summer literacy program among low-income children. Because surveying the large numbers of students in the program would require too many resources the library staff interviews 20 randomly chosen children among the low-income program attendees. The 20 sampled children are given a reading test before and after the program. a) (2 Points) Describe the population of this study? The population of this study is children who are from low-income families.b) (4 Points) The difference in the reading test scores (after – before) has mean 15 and standard deviation 5. Assuming the score differences are normally distributed, what percent of the children showed and improvement in reading ability?According to table of Standard Normal Distribution, P=(0.4978+0.5)= 99.87% of children will show improvement.c) (4 Points) What percent of children improved by more than 10 points? 1 SD z score of 1. According to table of Standard Normal Distribution P=(0.3413+0.500)= 84.13% improved more than 10 points.3.(12 points) The final total score (assignments + participation + exams + final project) on a Statistics course has mean of 90 and variance of 5. Assuming the final score X is normally distributed and given that the grading scheme is as follows: B-: 80 < X ≤ 83; B: 83< X ≤ 87; B+: 87 < X ≤ 90; A-: 90 < X ≤ 94; A: 94 < X ≤100;a) (4 points) What percentage of the students get an A, and A-, and a B+? s=√5=2.236 P(B+)=0.4099=40.99%P(A-)=4/2.236=1.78=z 0.4625= 46.25%P(A)=0.0375= 3.75%P(B+, A-, A)= 0.9099= 90.99%b) (4 points) How high need a score X be on this exam so that only 1% of the students score higher than X. What is this score called?1%= 0.0010= a z score of 3.09  3.09*2.23= 6.90924  90+6.90924= 96.91%This number is the 99 th percentile.c) (4 points) The final scores for another section of the course is also normally distributed with mean 90, but the variance is smaller than 5. How do the 95% percentiles compare between the two sections? Why?According to the table of Standard Normal Distribution, as the z-score gets larger, so does the percentage. Therefore, the section with a variance of 5 would have a greater range of scores in the 95% percentile because there is a greater z score, and thus a greater percentage.4. (10 points) Taylor et al. (1995) investigated the preferred ambient room temperature of young males. They found that the median preferred room temperature was 76.8 degrees Fahrenheit,with a standard deviation of 2.7 degrees. Assuming that the preferred temperatures are normallydistributed:a) (5 points) What proportion of young males would feel cold on a 65F room? (assume people feel cold if their preferred temperature is above the actual temperature)(65-76.8)/2.7=-4.37z-score= -4.37  Since the z-score is so large, it does not appear on the table and therefore the proportion of males that would feel cold in 65F is very unlikely and very slim.b) (5 points) On a hot summer day, what temperature do you need to set the AC so that 90% of young males have their preferred temperature above the room temperature?-1.208=(x-mean)/stdev-1.208=(x-76.8)/2.7-3.26=x-76.8x=73.54