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sveta [45]
2 years ago
10

SAT verbal scores are normally distributed with a mean of 430 and a standard deviation of 120 (based on data from the College Bo

ard ATP). (a) If a single student is randomly selected, find the probability that the sample mean is above 500. (b) If a sample of 35 students are selected randomly, find the probability that the sample mean is above 500. These two problems appear to be very similar. Which problem requires the application of the central limit theorem, and in what way does the solution process differ between the two problems?
Mathematics
1 answer:
Sav [38]2 years ago
6 0

Using the <u>normal distribution and the central limit theorem</u>, it is found that there is a:

a) 0.281 = 28.1% probability that the sample mean is above 500.

b) 0.0003 = 0.03% probability that the sample mean is above 500.

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

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 430, hence \mu = 430.
  • The standard deviation is of 120, hence \sigma = 120.

Item a:

The probability is the <u>p-value of Z when X = 500</u>, hence:

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

Z = \frac{500 - 430}{120}

Z = 0.58

Z = 0.58 has a p-value of 0.719.

1 - 0.719 = 0.281

0.281 = 28.1% probability that the sample mean is above 500.

Item b:

Sample of 35, hence n = 35, s = \frac{120}{\sqrt{35}}

Then:

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

By the Central Limit Theorem

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

Z = \frac{500 - 430}{\frac{120}{\sqrt{35}}}

Z = 3.45

Z = 3.45 has a p-value of 0.9997.

1 - 0.9997 = 0.0003

0.0003 = 0.03% probability that the sample mean is above 500.

To learn more about the <u>normal distribution and the central limit theorem</u>, you can take a look at brainly.com/question/24663213

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p = 2w + 2l              p = 42 
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You want to make either the length or width variable (I chose the width variable to be by itself. It doesn't matter) by itself by:

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Subtract both sides by either length or width depending on what variable you chose to make by itself (in this case I subtracted the length variable because I wanted the width variable by itself).
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You replace w with the equation above p/2 - l = w 
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Plug in all your numbers a = 110 and p = 42
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Move everything to the left side so l^{2} would be positive (makes the equation easier when l^{2} is positive).
l^{2} - 21l + 110 = 0

Factor.
(l -10)(l-11) = 0

Make each parenthesis set equal to 0.
l-11=0     l-10=0

Add.
l = 11  l = 10

By doing this you solve for both the width and length so the answer is:
w = 11    L = 10
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