Question

a. Ms. Newell gave her class a 10-question multiple-choice quiz with 4 choices. Let X = the number of questions that a randomly selected student in the class answered correctly. The mean and standard deviation of X are 7.6 and 1.32 respectively. To determine each student’s grade on the quiz, Ms. Newell will multiply his or her number of correct answers by 10. Let RV ‘G’ be the grade of a randomly chosen student in the class. i. Calculate the mean of RV ‘G’. Show your work. ii. Calculate the standard deviation of RV ‘G’. Show your work. ii. How are the mean and variance of G and X related? iii. If Johnny was simply guessing, what is the probability that he would pass? b. A report from Center for Health Statistics says that the height of a 20-year-old men when chosen at random, is a random variable (RV) X with mean height of 5.8 feet and standard deviation 0.24 feet. Height is "normally" distributed. i. Find the mean and standard deviation of the height in inches of a randomly selected 20-year-old men. (Note: There are 12 inches in a foot) ii. Calculate the 90th percentile for the height of 20 year old men. iii. Calculate the probability that the man is taller than 5' 11" tall.

Answer 1

a.) Mean(X) = 7.6

Sd(X) = 1.32

Given grade of a student :

G = 10 x ( X )

i) Mean of G = 10 x (7.6)

= 76.

ii) Standard deviation (X) = 10 x 1.32

= 13.2

Mean and variance of G is related to the mean of X with the same relation as G is related to X.

iii) let G(n) is minimum grade for passing or jonny has to answer atleast n question to pass.(n<=10)

probability of answering a question correctly = 1/4

probability of jonny to pass the exam = atleast n questions should be correclty answered

$\binom{10}{n}\times {1/4}^{n}\times {3/4}^{10-n} + \binom{10}{n+1}\times {1/4}^{n+1}\times {3/4}^{10-n-1}+ .........+ \binom{10}{10}\times {1/4}^{10}\times {3/4}^{10-10}$

b.)  

i) 1 foot= 12 inches
Mean height= 5.8 *12= 69.6 inches
Standard deviation= 0.24*12=2.88 inches

ii) To calculate the 90th percentile of given normal distribution = X = $\mu +Z\sigma$

where, $\mu$ = mean

  $\sigma$ = standard deviation

and ,  Z is the value from the standard normal distribution for the desired percentile.(Z for 90th percentile = 1.282)

  $\therefore$ X = 69.6 + 2.88 x 1.282

= 76.98432

iii) Probability = man's height / mean of the group

Man's height = 5 x 12 + 0.11 x 12 = 61.32

Probability= 61.32 / 69.6 =  0.8810