Question

question 4

1. A distribution has a standard deviation of a = 12 points. Find the 2-score for each of the following locations in a distribution by sketching a distribution (do not use an equation). (4 points) a. Above the mean by 4 points b. Below the mean by 6 points c. Below the mean by 18 points 2. A distribution has a standard deviation of a = 5 and u = 30. Find the score for each of the following locations in a distribution by using the Z-scale formula. (4 points) a. X = 40 b. X = 15 c. X = 22.5 8. Solve by drawing 3. Convert the following z scores to raw scores, given a u = 50 and distributions. Do not use a formula! (3 points) a. z= +1 b. Z=-.50 c. Z=- 2.0 4. In a population of exam scores, a score of X = 48 corresponds to a z=+1, while a score of X = 36 corresponds to corresponds to z= -50. Find the mean and standard deviation for the population by sketching a distribution and identifying the appropriate X and z scores. (Hint - draw it out! 3 points) A distribution with p = 60 and 6 = 12 is transformed to a standardized distribution with u = 50 and o = 10. Transform the scores from the original scale to the new scale. (Your answer should be an X value not a Z value; 3 points) a. X= 48 b. X = 66 c. X= 54 6. Find the proportion of a normal distribution that is: (4 points) a. Between z=-5 and z= +.5 b. Between z= -8 and z= +1.15 c. Greater than z = +1.85 d. Less than z = +1.25 7. Information from the Department of Motor Vehicles indicates that the average age of licensed drivers is = 45.7, and the standard deviation is o = 12.5 years. Assuming that the distribution is normal, answer the following questions. (10 points) a. What proportion of licensed drivers are older than 50? b. What proportion are younger than 50? c. What is the median age? d. What age separates the top 20% of drivers from the bottom 80%? e. Young drivers are considered the most dangerous because they are less experienced. What proportion of drivers are between the ages of 18 and 25 - aka the "high risk" group?

Answer 1

4 let mean ari standard deviation o, far aalei X= 48 h=x- for =-0.5 x= 36 > X-M=zo Ź X-M= Zo → 48-M= o. Ə 36-M = -0.50 → M to 748 (1) => N-0.50 =36-6) solve (0) and (2), and get, Mto -48 from () weget, M -0.50 -36 pl = 48-o = 48-8-40 1050 = 12 os -8 1.5 Mean 140 standaod deviation o-18 36124048 -0.5MEO 1