1) Consider a random variable X that is normally distributed with a mean and standard deviation ?. Converting X to a standard normal distribution _____.

creates a new random variable called Z which is normally distributed with a mean of zero and an unknown standard deviation.

creates a new random variable called Z with a mean of zero with an unknown standard deviation and unknown distribution.

creates a new random variable called Z which is normally distributed with a mean of zero and standard deviation of 1.

creates a new random variable called Z which is normally distributed with an unknown mean and a standard deviation of 1.

2)

Question 2 The Z-transformation can be written as Z = Q What is x? LT 0 Unknown 0 An observation {realization} of the random variable K O The mean of a sample of observations from the random variable X 0 The standard deviation of the population Question 3 The Z-transformation can be written as Z = Q What is x _# E" 0 Unknown 0 An observation of the random variable X C) The distance from the observed value, K, and the mean of the random variable K. O The distance from the observed value, K, and the standard deviation of the random variable X. Question 4 The Z-transformation can be written as z = Q. What is o? 0 Unknown 0 An observation of the random variable X 0 The mean of the random variable X. 0 The standard deviation of the random variable X. Question 5 The Z-transformation can be written as z = Q. How might we interpret this ratio? 0 The number of standard deviations x is from the mean of X. 0 Az-score which is a realization of the random variable Z. O The number of times the standard deviation goes into the distance an observation of the random variable X lies from the mean of X 0 All of the above Question 6 What is the probability of the complement of A under this standard normal curve equals (A is the area to the right of the mean)? Mean= 0 A 2 0 0 O 1 O 0.75 O 0.5Question 8 The area A under this standard normal curve equals _?_. Mean = 0 Z =1.5 O The same as area B due to symmetry O 0.5 minus the area B O 1 minus the area B O The area A minus B plus the area the left of 0Question 9 To calculate the shaded area A under this standard normal curve __?__ "HIRE za 2; {3- Find the area from the mean to 22 and add that to Area B {3- Find the area from the mean to 22 and add that to the area from the mean to 21 :1 Find The area from the mean to 22 C- Add the area of A and B and subtract that from I15 Question 10 How would we find the shaded area under this standard normal curve in Excel? Mean = 0 Z1 O 1 - NORM.S.DIST(Z1,1) O NORM.S.DIST(Z1,1) O NORM.S.DIST(Z1,1) - NORM.S.DIST(0,1) O NORM.S.DIST(0,1) - NORM.S.DIST(Z1,1)Suppose you knew that the heights of adult females in the US is normally distributed as in the following table. Table 1 Mean = 63.7 inches Standard Deviation = 2.2 inches If you wanted to find the probability that an adult female chosen at random would have a height of 70 inches or more, how would you calculate the Z-score to covert this to a standard normal distribution? O z = 12-60 2.5 O z = 10-63.7 2.2 O z = $17-70 2.2 Question 14 1 pt Using Table 1, what is the probability that an adult female chosen at random would have a height of 72 inches or more? O 0.0001 or about 0.01% 0 0.9772 or about 98% O 0.0548 or about 5.5% O 1.477 or about 148%Question 15 Using Table 1, what is the probability that an adult female chosen at random would have a height less than 60 inches? O 0.3446 or about 34.5% O 0.0668 or about 7% O 0.0465or about 4.7% O 1.477 or about 148% Question 16 Using Table 1, what is the probability that an adult female chosen at random would have a height between 60 and 65 inches? O 0.0228 or about 2.3% O 0.3413 or about 34% O 0.6759 or about 68% O 0.6006 or about 60% Question 17 Using Table 1, what is the probability that an adult female chosen at random would have a height between 68 and 69 inches? O 0.0228 or about 2.3% O 0.1554 or about 15.5% O 0.1499 or about 15% O 0.3413 or about 34%Set up 1: Let the random variable X represent the profit made on a randomly selected day by the Ben Franklin Five and Dime store on 185th Street. Assume X is Normal with a mean of $425 and standard deviation $50. Pr(X > $450) equals O 0.2525. O 0.2119. O 0.3085. O 0.8000.Question 19 Using Set up 1, the probability is 0.75 that on a randomly selected day the store's profit will be less than _?__. O $458.72. O $437.67. O $372.67. O $381.40. Question 20 Using Set up 1, if four days are selected at random, the probability that profit will exceed $450 on each of those days is: (Hint: don't forget the rules of probability.) O 0.0091 O 0.0041 O 0.2119 O 0.8476