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
A First Course in Probability (10th Edition)
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ISBN:9780134753119
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Chapter1: Combinatorial Analysis
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Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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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
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**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 \( X \).
\[ X = 36,000 \]
(b) Using the software tool above, find the probability stated in part (a).
\[
\mu = 38,500, \, \sigma = 2,500, \, n = 60
\]
\[
z = \frac{X - \mu}{\sigma/\sqrt{n}} = \frac{36,000 - 38,500}{2500/\sqrt{60}} = -119.1763
\]
\[
Pr(\bar{X} \leq 36,000) = Pr(Z \leq \frac{36,000 - 38,500}{2500/\sqrt{60}}) = Pr(Z \leq -119.1763) = 0
\]
(c) Using the software tool above, graph the probability of the statement in part (b).
### Diagram Explanation:
- The graph shows the probability \( Pr(X \leq 36,000) \) on the vertical axis.
- The horizontal axis represents the tire mileages.
- The plot appears to demonstrate a normal distribution curve, highlighting the area corresponding to the probability of a sample mean being less than 36,000 miles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb27aec09-4413-42f4-8ddd-e0b51dd64302%2F9e91fe81-cad2-49f0-becd-a157454a8814%2Fl30yxff_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:** Arianna Pierce
---
In this lab, we cover Normal Probability cases based on a sample and not on the entire population. These cases all require an adjustment to the standard deviation.
**Note:** Students should use the tool link: [Normal Probability Calculator - Sampling Distributions](https://mathcracker.com/normal-probability-calculator-sampling-distributions)
---
**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 \( X \).
\[ X = 36,000 \]
(b) Using the software tool above, find the probability stated in part (a).
\[
\mu = 38,500, \, \sigma = 2,500, \, n = 60
\]
\[
z = \frac{X - \mu}{\sigma/\sqrt{n}} = \frac{36,000 - 38,500}{2500/\sqrt{60}} = -119.1763
\]
\[
Pr(\bar{X} \leq 36,000) = Pr(Z \leq \frac{36,000 - 38,500}{2500/\sqrt{60}}) = Pr(Z \leq -119.1763) = 0
\]
(c) Using the software tool above, graph the probability of the statement in part (b).
### Diagram Explanation:
- The graph shows the probability \( Pr(X \leq 36,000) \) on the vertical axis.
- The horizontal axis represents the tire mileages.
- The plot appears to demonstrate a normal distribution curve, highlighting the area corresponding to the probability of a sample mean being less than 36,000 miles.
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