Question

a. Time to finish a quiz in minutes follows a normal distribution with mean 6 and standard deviation .7.  A student took 4.8 minutes. What is this student’s z-score?

b. Time to finish a quiz in minutes follows a normal distribution with mean 6 and standard deviation .7. What proportion of students complete the quiz in less than 5.2 minutes? Use the normalcdf command on your calculator with 6 as the mean, .7 as the standard deviation, a very large negative number as the lower bound, and 5.2 as the upper bound. normalcdf is on the dist menu of your calculator. If you did not bring your calculator, you can skip down to where Excel instructions are given in the lab, and come back to do these problems using Excel.

c. Time to finish a quiz in minutes follows a normal distribution with mean 6 and standard deviation .7.  What proportion of students took longer than 7.2 minutes to complete the quiz? Use the normalcdf command on your calculator with 6 as the mean, .7 as the standard deviation, 7.2 as the lower bound and a large positive number as the upper bound.

d. Time to finish a quiz in minutes follows a normal distribution with mean 6 and standard deviation .7. What proportion of students took between 4.6 and 7 minutes? Use the normalcdf command on your calculator with 6 as the mean, .7 as the standard deviation, 4.6 as the lower bound, and 7 as the upper bound.

e. Find the first quartile  and third quartile  of the Normal(6, .7) distribution. Give your answers to four decimal places. Use the invNormal command on your calculator with 6 as the mean, .7 as the standard deviation and the appropriate left tail areas for the quartiles.

Answer 1

sol o = time to finish a quiz XNNl M= 6 = 0.7) a) z-score= x-ll For x = 408 Z- score = 9.8 - 6 0.7 ?????????????? Z- score= -1.7143 bp( x.5.2] = P(x-4 < 5:2-8 ] = P[ 22-101429) = 0.1265 c) P(x > 7.2] = P( XH > 7. 2. Of] - P[ Z> 1.4143] Z > 1.7143 = 0.0432

d) P[ 4.6 L XL7] = Plyob b < x-M L 7 - 1 1 2 07 - P[-2.22 1.42867 = 0.900 e) eis first quartile if P{ x < c] = 0.25 PL x L C. b ] = 0.25 P[ Z_-0674] = 0.25 >C-6 = - 0.674 Oo7 - C= 5.5282 is first quartile

it is third quantile of Pľ x <t] = 0.75 P[ X-4 < t-o] = 0.75 P[ zato = 0.75 • P[ z < 0.674] = 0.45 - o tb = 0.674 = t = 6.4718 is third quartile