AnimeBrainly
Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

The average breaking strength of a certain brand of steel cable is 2000 pounds, with a standard deviation of 100 pounds. A sample of 20 cables is selected and tested. Find the sample mean that will cut off the upper 95% of all samples of size 20 taken from the population. Assume the variable is normally distributed.

Sagot :

Answer:

1963.2 pounds (lbs.)

Step-by-step explanation:

Things to understand before solving:

  • - Normal Probability Distribution
  1. The z-score formula can be used to solve normal distribution problems. In a set with mean ц and standard deviation б, the z-score of a measure X is given by: [tex]Z=\frac{X-u}{a}[/tex]

The Z-score reflects how far the measure deviates from the mean. After determining the Z-score, we examine the z-score table to determine the p-value associated with this z-score. This p-value represents the likelihood that the measure's value is less than X, or the percentile of X. Subtracting 1 from the p-value yields the likelihood that the measure's value is larger than X.

  • - Central Limit Theorem
  1. The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean ц and standard deviation б , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean ц  and standard deviation [tex]s=\frac{a}{\sqrt{n} }[/tex]

As long as n is more than 30, the Central Limit Theorem may be applied to a skewed variable. A specific kind of steel cable has an average breaking strength of 2000 pounds, with a standard variation of 100 pounds.

This means, ц  = 2000 and б = 100.

A random sample of 20 cables is chosen and tested.

This means that n = 20, [tex]s=\frac{100}{\sqrt{120} } =22.361[/tex]

Determine the sample mean that will exclude the top 95 percent of all size 20 samples drawn from the population.

This is the 100-95th percentile, or X when Z has a p-value of 0.05, or X when Z = -1.645. So [tex]Z=\frac{X-u}{a}[/tex]

  • By the Central Limit Theorem

[tex]Z=\frac{X-u}{a} \\-1.645=\frac{X-2000}{22.361} \\X-2000=-1.645*22.361[/tex]

[tex]X =1963.2[/tex]

Answer:

The sample mean that will cut off the top 95% of all size 20 samples obtained from the population is 1963.2 pounds.

Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.