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.
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.
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Section10.3: Measures Of Spread
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Binomial Distribution
Binomial is an algebraic expression of the sum or the difference of two terms. Before knowing about binomial distribution, we must know about the binomial theorem.
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Question
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
![LAB-4 Chapter 7
Print the lab and write or type the answers in the space provided
Student Full Name: Hrzanna Prece
Sample size=6
Sample mean= 38,500
Standard deviation
こ2,SoO
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.
=39500,0Ea500,n6Q
X<36,000 ?
(b) Using the software tool above, find the probability stated on part (a) r
= 38,500 ,G=005, N= 60
X-H=3G000 - 38500 -119,1763
2500NG
こ。
36,800 - 36,500 - Pr(zS 119.1763)=0
Prlås36,000) - Pr(zs
ఇకం
(e) Using the software tool above, graph the probability of stated on part (b) Pr (xe 36,0000
G.olg
G.016
0.008.
0.0004
0.0000+
500
lopoo 15,0° 29000 25GOO 30,000 35,000 400](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb27aec09-4413-42f4-8ddd-e0b51dd64302%2F03d4c9d4-de28-403d-9a85-1838008b4de3%2F7w3gf0f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:LAB-4 Chapter 7
Print the lab and write or type the answers in the space provided
Student Full Name: Hrzanna Prece
Sample size=6
Sample mean= 38,500
Standard deviation
こ2,SoO
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.
=39500,0Ea500,n6Q
X<36,000 ?
(b) Using the software tool above, find the probability stated on part (a) r
= 38,500 ,G=005, N= 60
X-H=3G000 - 38500 -119,1763
2500NG
こ。
36,800 - 36,500 - Pr(zS 119.1763)=0
Prlås36,000) - Pr(zs
ఇకం
(e) Using the software tool above, graph the probability of stated on part (b) Pr (xe 36,0000
G.olg
G.016
0.008.
0.0004
0.0000+
500
lopoo 15,0° 29000 25GOO 30,000 35,000 400
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