Question

(15.48 S-AQ) The scores of 12th-grade students on the National Assessment of Educational Progress year 2000 mathematics test have a distribution that is approximately Normal with mean µ = 298 and standard deviation s = 34.

1. Choose one 12th-grader at random. What is the probability (± ± 0.1) that his or her score is higher than 298? Higher than 332 (± ± 0.001)?

2. Now choose an SRS of 16 twelfth-graders and calculate their mean score x⎯⎯⎯ x ¯ . If you did this many times, what would be the mean of all the x⎯⎯⎯ x ¯ -values?

3. What would be the standard deviation (± ± 0.1) of all the x⎯⎯⎯ x ¯ -values?

4. What is the probability that the mean score for your SRS is higher than 298? (± ± 0.1) Higher than 332? (± ± 0.0001)

Answer 1

X: The score of 12th-grade students on the National Assessment of Educational Progress year 2000 mathematics test

X follows Normal distribution with mean =298 and standard deviation =34

1.

Probability (± ± 0.1) that his or her score is higher than 298 = P(X>298) = P(X>)

for Normal distribution P(X) = P(X>) = 0.5

By z-score method,

P(X>298) = 1-P(Z298)

Z-score for 298 = (298-298)/34 = 0

From standard normal tables , P(Z0) =0.5

P(Z298) = P(Z0) =0.5

P(X>298) = 1-P(Z298) = 1-0.5 =0.5

Probability that his or her score is higher than 298 =0.5

Probability that his or her score is Higher than 332 =P(X>332)

Mean + 1 standard deviation = 298+34=332

The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean. The empirical rule can be broken down into three parts:

• 68% of data falls within the first standard deviation from the mean.
• 95% fall within two standard deviations.
• 99.7% fall within three standard deviations.

As depicted in the diagram,

P(X>Mean + 1 standard deviation) = 16% i.e 0.16

Probability that his or her score is Higher than 332 =P(X>332) =0.16;

By using z-score :

P(X>332) = 1-P(Z332)

Z-score for 332 = (332-298)/34 = 34/34 = 1

From standard normal tables : P(Z1) = 0.8413

P(Z332) = P(Z1) = 0.8413

P(X>332) = 1-P(Z332) = 1 - 0.8413 = 0.1587

Probability that his or her score is Higher than 332 =0.1587

2. Sample size : n=16

By central limit theorem,

the mean of all the x :

3.

The standard deviation of all the x :

4.

By central limit theorem,

Sampling distribution sample mean : : i.e mean score for your SRS follows normal distribution with mean 298 and standard deviation 8.5

probability that the mean score for your SRS is higher than 298 = P( > 298) = P( > ) =0.5

By z-score method ; P( > 298) = 1-P(298)

z-score for 298 = (298-298)/8.5 = 0

From standard normal tables P(Z0) = 0.5

P(298) = P(Z0) = 0.5

P( > 298) = 1-P(298) = 1-0.5 =0.5

probability that the mean score for your SRS is higher than 298 = 0.5

Probability that the mean score for your SRS higher than 332 = P( > 332)

P( > 332) = 1-P( 332)

Z-score for 332 = (332-298)/8.5 = 4

Generally normal tables are available for +3 to -3 ; anything beyond 3: P(Z3) is considered to be equal to 1.

Using excel , P(Z4) = 0.999968329

P( 332) = P(Z4) = 0.999968329

P( > 332) = 1-P( 332) = 1-0.999968329 = 0.000031671

Probability that the mean score for your SRS higher than 332 = 0.000031671

Probability that the mean score for your SRS higher than 332 = 0.0000 for (± 0.0001)