Question

Chapter 7 Test Due in 7 hours, 28 minutes. Due Thu 07/30/2020 11:59 pm Questions Question 1 (0/10) Question 2 [10/10) Question 3 (0/10) Question 4 [10/10) Question 5 [10/10) Question 6 (0/10) Question 7 [10/10) Question 8 (0/10) Suppose that the amount of time that students spend studying in the library in one sitting is normally distributed with mean 45 minutes and standard deviation 21 minutes. A researcher observed 6 students who entered the library to study. Round all answers to 4 decimal places where possible a. What is the distribution of X? X-N b. What is the distribution of I? EN c. What is the distribution of r-N d. If one randomly selected student is timed, find the probability that this student's time will be between 32 and 39 minutes. e. For the 6 students, find the probability that their average time studying is between 32 and 39 minutes. Grade: 40/80 f. Find the probability that the randomly selected 6 students will have a total study time more than 264 minutes g. For part e) and f), is the assumption of normal necessary? Yes No 1. The top 20% of the total study time for groups of 6 students will be given a sticker that says "Great dedication". What is the least total time that a group can study and still receive a sticker? minutes Hint:

Answer 1

a)X ~N(45,21)

b)

std error=σx̅=σ/√n=

8.5732

xbar ~N(45,8.5732)

c)

total expected value =45*6=270

standard deviation =21*√6 =51.4393

Σx ~N(270 , 51.4393)

d)

probability =P(32<X<39)=P((32-45)/21)<Z<(39-45)/21)=P(-0.62<Z<-0.29)=0.3875-0.2679=0.1196

e)

probability =P(32<X<39)=P((32-45)/8.573)<Z<(39-45)/8.573)=P(-1.52<Z<-0.7)=0.242-0.0647=0.1773

f)

probability =P(X>264)=P(Z>(264-270)/51.439)=P(Z>-0.12)=1-P(Z<-0.12)=1-0.45357=0.5464

g) yes (since n<30)

h)

for 80th percentile critical value of z=0.84162123

therefore corresponding value=mean+z*std deviation=313.2924