Mathematics, 23.06.2021 19:20 shanewugden7362

Suppose that attendance at the concerts by the band "Keane" is a normally distributed random variable X with a mean of 18,500. You are told that P(X ≥ 15,000) = 0.6981. What are the two values of X that delineate the "82% middle pack" of this random variable? A random variable has a population mean equal to 1,973 and population variance equal to 892,021. Your interest lies in estimating the population mean of this random variable. With that in mind, you take a representative sample of size 79 from the population of the random variable. You then use this sample data to calculate the sample average as an estimate for the population mean.

Required:
Using your knowledge about the central limit theorem (CLT), and assuming that the CLT has already "established itself" / "kicked in" when the sample size is 79, what is the probability that the sample average that you calculated will lie between 1,702 and 1,948?

Answers: 2

Show answers

Answers

Answer from: adelfaali99

The two values of X that delineate the "82% middle pack" of this random variable are 9480 and 27520.

0.4017 = 40.17% probability that the sample average that you calculated will lie between 1,702 and 1,948.

Step-by-step explanation:

To solve the first question, we use the normal distribution, while for the second quetion, it is used with the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

First question:

Mean of 18,500:

This means that $\mu = 18500$

You are told that P(X ≥ 15,000) = 0.6981.

This means that when X = 15000 , Z has a o-value of 1 - 0.6981 = 0.3019, which means that when X = 15000, Z = -0.52 . We use this to find $\sigma$ . So

$Z = \frac{X - \mu}{\sigma}$

$-0.52 = \frac{15000 - 18500}{\sigma}$

$0.52\sigma = 3500$

$\sigma = \frac{3500}{0.52}$

$\sigma = 6731$

What are the two values of X that delineate the "82% middle pack" of this random variable?

Between the 50 - (82/2) = 9th percentile and the 50 + (82/2) = 91st percentile.

9th percentile:

X when Z has a p-value of 0.09, so X when Z = -1.34.

$Z = \frac{X - \mu}{\sigma}$

$-1.34 = \frac{X - 18500}{6731}$

X - 18500 = -1.34*6731

X = 9480

91st percentile:

X when Z has a p-value of 0.91, so X when Z = 1.34.

$Z = \frac{X - \mu}{\sigma}$

$1.34 = \frac{X - 18500}{6731}$

X - 18500 = 1.34*6731

X = 27520

The two values of X that delineate the "82% middle pack" of this random variable are 9480 and 27520.

Question 2:

A random variable has a population mean equal to 1,973 and population variance equal to 892,021.

This means that $\mu = 1973, \sigma = \sqrt{892021} = 944.5$

Sample of 79:

This means that $n = 79, s = \frac{944.5}{\sqrt{79}}$

What is the probability that the sample average that you calculated will lie between 1,702 and 1,948?

This is the p-value of Z when X = 1948 subtracted by the p-value of Z when X = 1702. So

X = 1948

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1948 - 1973}{\frac{944.5}{\sqrt{79}}}$

Z = -0.235

Z = -0.235 has a p-value of 0.4071

X = 1702

$Z = \frac{X - \mu}{s}$

$Z = \frac{1702 - 1973}{\frac{944.5}{\sqrt{79}}}$

Z = -2.55

Z = -2.55 has a p-value of 0.0054

0.4071 - 0.0054 = 0.4017

0.4017 = 40.17% probability that the sample average that you calculated will lie between 1,702 and 1,948.

Answer from: Quest

Ithink its 126 hope its
If one angle of a set of vertical angles measure 63 the sum of the vertical angles is 360 12618090

Answer from: Quest

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.

Another question on Mathematics

Mathematics, 21.06.2019 19:00

Tabby sells clothing at a retail store, and she earns a commission of 6.25% on all items she sells. last week she sold a pair of jeans for $32.50, a skirt for $18.95, and 2 blouses for $14.95 each. what was tabby's total commission? a $4.15 b $5.08 c $81.35 d $348.60

Answers: 1

Answer

Mathematics, 21.06.2019 20:30

Secant ac and bd intersect at point e inside f what is the measure of aed if measurements cd equals 30 a b is 50 and cb is 170

Answers: 1

Answer

Mathematics, 21.06.2019 22:20

As voters exit the polls, you ask a representative random sample of voters if they voted for a proposition. if the true percentage of voters who vote for the proposition is 63%, what is the probability that, in your sample, exactly 5 do not voted for the proposition before 2 voted for the proposition? the probability is

Answers: 2

Answer

Mathematics, 22.06.2019 00:30

Carl is hanging pictures in his home. there is a space on a wall that has a width of 4 feet. carl has pictures with different widths. which picture width would fit in the space on the wall? a. 46 inches b. 60 inches c. 50 inches d. 54 inches

Answers: 1

Answer

You know the right answer?

Suppose that attendance at the concerts by the band "Keane" is a normally distributed random variabl...

Questions