LAB-4 Chapter 7 Sample stze= 60 Sample means 38,500 Stardard deViahon =2,S00 Print the lab and write or type the answers in the space provided sudent Full Name: Arranna Precen Jn this lab we cover Normal Probability cases based on a sample and not on the entire population, These cases all require and adjustment to the standard deviation. Note: Students should use the tool link: https://mathcracker.com/normal-probability-calculator-sampling-distributions (Include the work from the software as part of your answer before submitting the lab) 1. A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. The manufacturer tests 60 such tires. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal. (a) Let X = number of miles on a single tire. Write the question above in terms of this variable X. u= 39500 02500, ne 60 メニ36,000 (b) Using the software tool above, find the probability stated on part (a) H= 36,500 , G- B005, n=60 %3D =3G000 - 38500 l19,1763 2500/6 ここ %3D Priis 36,000) - Pr{zs 36,00-36,500 -Pr(zs 119.1763)-0 2500/160 =Rr(zS 119.1763)-G c) Using the software tool above, graph the probability of stated on part (b) Pr(xe 36,000 3.016 9.014 0.008
LAB-4 Chapter 7 Sample stze= 60 Sample means 38,500 Stardard deViahon =2,S00 Print the lab and write or type the answers in the space provided sudent Full Name: Arranna Precen Jn this lab we cover Normal Probability cases based on a sample and not on the entire population, These cases all require and adjustment to the standard deviation. Note: Students should use the tool link: https://mathcracker.com/normal-probability-calculator-sampling-distributions (Include the work from the software as part of your answer before submitting the lab) 1. A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. The manufacturer tests 60 such tires. On the assumption that the actual population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal. (a) Let X = number of miles on a single tire. Write the question above in terms of this variable X. u= 39500 02500, ne 60 メニ36,000 (b) Using the software tool above, find the probability stated on part (a) H= 36,500 , G- B005, n=60 %3D =3G000 - 38500 l19,1763 2500/6 ここ %3D Priis 36,000) - Pr{zs 36,00-36,500 -Pr(zs 119.1763)-0 2500/160 =Rr(zS 119.1763)-G c) Using the software tool above, graph the probability of stated on part (b) Pr(xe 36,000 3.016 9.014 0.008
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section: Chapter Questions
Problem 25SGR
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Question
B) using the software tool above, what is the probability stated on part (a)
C) use the software tool above, graph the probability of stated on part b
Transcribed Image Text:LAB-4 Chapter 7
Sample sIze= 60
Sample meanz 38,500
Stardard deViarion =2,Sa0
Print the lab and write or type the answers in the space provided
Sudent Full Name: Hrranna Prece
In this lab we cover Normal Probability cases based on a sample and not on the entire population. These cases
all require and adjustment to the standard deviation.
Note: Students should use the tool link: https://mathcracker.com/normal-probability-calculator-sampling-distributions
(Include the work from the software as part of your answer before submitting the lab)
1. A prototype automotive tire has a design life of 38,500 miles with a standard deviation of 2,500 miles. The
manufacturer tests 60 such tires. On the assumption that the actual population mean is 38,500 miles and the
actual population standard deviation is 2,500 miles, find the probability that the sample mean will be less
than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal.
(a) Let X = number of miles on a single tire. Write the question above in terms of this variable X.
=385000 2500,n6
X<36,000 ?
(b) Using the software tool above, find the probability stated on part (a)
H= 36,500 , G=25005, n=6o
%3D
x-N=3G00o - 38500 -119.l763
25001
ここ
Pr(スs36,000) - Pr(Z< 300-3,500 =Prz<1R.17630
2500/160
c) Using the software tool above, graph the probability of stated on part (b) r (xe 36,00G
G.016
0.008
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