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photoshop1234
4 days ago
11

Sales prices of baseball cards from the 1960's are known to possess a skewed-right distribution with a mean sale price of $5.25

and a standard deviation of $2.80. suppose a random sample of 100 cards from the 1960's is selected. describe the sampling distribution for the sample mean sale price of the selected cards.
Mathematics
2 answers:
zzz [11.9K]4 days ago
6 0

The mean will be $5.25 and the standard error calculates to $0.28

Based on the details given,

  • The sales prices of baseball cards from the 1960s =?
  • The mean sale price stands at $5.25
  • The standard deviation is $2.80

To answer your inquiry if a sample of 100 cards from 1960 is randomly selected, it results in a normal distribution with a mean of $5.25 and a standard error of $0.28.

Expounding further

as the standard deviation (σ) is established to be $2.80

Consequently, the formula for the standard error becomes SD divided by the square root of n

upon substitution of values, we find:

2.80 divided by the square root of 100

= 0.28.

Thus, the calculated standard error is $0.28.

Utilizing the central limit theorem assists in determining the mean, which suggests that the sample mean's distribution is close to normal, holding its mean and standard error.

So, the mean remains $5.25

When considering a sample of size Y from a specific population, one can compute statistics like proportion, mean, and standard deviation from the possible samples.

Therefore, the probability distribution of the calculated statistics (mean, proportion, and standard deviation) is termed the sampling distribution, while its standard error also equates to standard deviation.

However, the variability of a sampling distribution can be determined by its standard deviation based on three main factors:

  • The total number of observations in the population = N
  • The total number of observations in the sample = n
  • The method of selecting the random samples

Thus, the average will be $5.25 and the standard error will be $0.28

LEARN MORE:

  • At Appalachian State University, it's noted that 52% of undergraduates are female
  • Sales prices of baseball cards from the 1960s reveal a skewed-right distribution with a mean sale price of $5.25 and a standard deviation of $2.80

KEYWORDS:

  • standard deviation
  • sample mean
  • skewed right distribution
  • selected cards
  • sample distribution
Zina [12K]4 days ago
3 0

Provided the details regarding the sale prices of baseball cards from the 1960s:

It's established that they display a skewed-right distribution

The average sale price is $5.25

The standard deviation measures $2.80.

 

If we draw a random sample of 100 cards from that era:

The distribution will be Normal

The mean will remain at $5.25

The standard error turns out to be $0.28

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I believe it takes around 11 hours.

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babunello [11366]

The diagram is not available, so I included a supplementary figure.

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The segments ST and UT are equal in length.

Detailed explanation:

As seen in the additional figure

∵ ST and UT act as tangents to circle K at points S and U respectively

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Thus, UT ⊥ KU at point U

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1 month ago
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Which expression is equivalent to (3m^-2n)^-3/6mn^-2? Assume m=0, n=0.
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This is the solution:

(3m^-2 n)^-3 / 6mn^-2
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Answer:

Let's explore this step-by-step:

Greetings!

Here are the variables

X: daily cost of hotel rooms

Y: expenses for entertainment

Refer to the second attachment for the scatter plot.

The population regression equation can be expressed as E(Yi)= α + βXi

To find the y-intercept and the slope of the regression line, apply the following formulas:

a= Y[bar]-bX[bar]b= \frac{sum XY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }

In this scenario, n= 9; ∑X= 945; ∑X²= 103325; ∑Y= 1134; ∑Y²= 148804; ∑XY= 123307

Calculating X[bar]= ∑X/n= 945/9= 105

Calculating Y[bar]= ∑Y/n= 1134/9= 126

Thus, it follows that a= 126 - 1.03*105= 17.49b= \frac{123307-\frac{945*1134}{9} }{103325-\frac{(945)^2}{9} }= 1.03

The regression equation is then given by ^Y= 17.49 + 1.03Xi

Slope interpretation: For each unit increase in the daily hotel rate, the expected average spending on entertainment rises by $1.03.

Given that the room rate in Chicago is $128 (X), we can predict the entertainment expenditure (Y) using the estimated regression line:

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The anticipated spending on entertainment for Chicago is $149.33

I trust this will assist you!

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Answer:

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By simplifying, we can derive:

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Thus, the minimum sample size required is 315.

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