Mathematics, 21.11.2020 22:30 allieballey0727

Given a test that is normally distributed with a mean of 100 and a standard deviation of 12, find: (a) the probability that a single score drawn at random will be greater than 110

(b) the probability that a sample of 25 scores will have a mean greater than 105

(c) the probability that a sample of 64 scores will have a mean greater than 105

(d) the probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

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Answer from: han68

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

a) 0.2033 = 20.33% probability that a single score drawn at random will be greater than 110.

b) 0.0188 = 1.88% probability that a sample of 25 scores will have a mean greater than 105.

c) 0.0004 = 0.04% probability that a sample of 64 scores will have a mean greater than 105.

d) 0.095 = 9.5% probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105.

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:

Mean of 100, hence $\mu = 100$ .Standard deviation of 12, hence $\sigma = 12$ .

Item a:

This probability is 1 subtracted by the p-value of Z when X = 110, hence:

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

$Z = \frac{110 - 100}{12}$

Z = 0.83

Z = 0.83 has a p-value of 0.7967.

1 - 0.7967 = 0.2033

0.2033 = 20.33% probability that a single score drawn at random will be greater than 110.

Item b:

Sample of 25, hence $n = 25, s = \frac{12}{\sqrt{25}} = 2.4$ .

This probability is 1 subtracted by the p-value of Z when X = 105, hence:

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

By the Central Limit Theorem

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

$Z = \frac{105 - 100}{2.4}$

Z = 2.08

Z = 2.08 has a p-value of 0.9812.

1 - 0.9812 = 0.0188

0.0188 = 1.88% probability that a sample of 25 scores will have a mean greater than 105.

Item c:

Sample of 64, hence $n = 64, s = \frac{12}{\sqrt{64}} = 1.5$ .

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

$Z = \frac{105 - 100}{1.5}$

Z = 3.33

Z = 3.33 has a p-value of 0.9996.

1 - 0.9996 = 0.0004

0.0004 = 0.04% probability that a sample of 64 scores will have a mean greater than 105.

Item d:

Sample of 16, hence $n = 16, s = \frac{12}{\sqrt{16}} = 3$ .

Both 105 and 95 are the same distance of the mean, so we find one probability, and multiply by 2.

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

$Z = \frac{105 - 100}{3}$

Z = 1.67

Z = 1.67 has a p-value of 0.9525.

1 - 0.9525 = 0.0475.

2 x 0.0475 = 0.095

0.095 = 9.5% probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105.

A similar problem is given at link

Answer from: tabbycat2574

(a) 0.2033

(b) 0.0188

(d) 0.095

Step-by-step explanation:

(a) the probability that a score at random is greater than 110 is obtained with a normal distribution of mean 100 and standard deviation 12 can be estimated using the z-table for Z = (110 - 100)/12 = 0.83

So P (X > 110) = P (Z > 0.83) = 0.2033

(b) Probability that a sample of 25 scores will have a mean greater than 105:

we use a standard distribution with the same mean (100) but the standard distribution reduced by a factor of $\sqrt{25} = 5$ . That is a standard deviation of 12/5 = 2.4. which gives a Z-value of (105-100) / 2.4 = 2.08

P (X> 105) = P (Z > 2.08) = 0.0188

we use a standard distribution with the same mean (100) but the standard deviation reduced by a factor of $\sqrt{64} = 8$ . That is a standard deviation of 12/8 = 1.5. which gives a Z-value of (105-100) / 1.5 = 3.33

P (X> 105) = P (Z > 3.33) = 0.0004

(d) the probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

This will be the addition of the two probabilities. We use a standard distribution with the same mean (100) but the standard distribution reduced by a factor of $\sqrt{16} = 4$ . That is a standard deviation of 12/4 = 3. which gives us two different Z values to study:

(105-100) / 3 = 1.67

and for X= 95 ==> Z = (95 - 100)/3 = - 1.67

P (X > 105) = P (Z > 1.67) = 0.0475

P (X < 95) = P (Z < -1.67) = 0.0475

which add up to: 0.095.

Answer from: Quest

The answer would be b

Answer from: Quest

2x^2+5x+17

your welcome. : )

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