Question

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.

Answer 1

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: $Z=\frac{X-u}{a}$

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 $s=\frac{a}{\sqrt{n} }$

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, $s=\frac{100}{\sqrt{120} } =22.361$

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 $Z=\frac{X-u}{a}$

• By the Central Limit Theorem

$Z=\frac{X-u}{a} \\-1.645=\frac{X-2000}{22.361} \\X-2000=-1.645*22.361$

$X =1963.2$

Answer:

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