The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009):

Assume that the population standard deviation on each part of the test is = 100.

a. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test (to 4 decimals)?

b. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test (to 4 decimals)?

c. What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)?

a. P(492 < x-bar < 512)

z = (492-502)/100/√90

z = -0.95 is 0.1711

z = (512-502)/100/√90

z = 0.95 is 0.8289

P(-0.95< z < 0.95) = 0.6578

b. P(505 < x-bar < 525)

z = (505-515)/100/ √90

z = -0.95

z = (525-515)/100/ √90

z = 0.95

P(-0.95< z < 0.95) = 0.6578

c. P(484 < x-bar < 504)

z = (484-494)/100/√100

z = -1 is 0.1587

z = (504-494)/100/√100

z = 1 is 0.8413

P(-1< z <1) = 0.6826

a. P(492 < x-bar < 512)

z = (492-502)/100/√90

z = -0.95 is 0.1711

z = (512-502)/100/√90

z = 0.95 is 0.8289

b. P(505 < x-bar < 525)

z = (505-515)/100/ √90

z = -0.95

z = (525-515)/100/ √90

z = 0.95

P(-0.95< z < 0.95) = 0.6578

P(-0.95< z < 0.95) = 0.6578

c. P(484 < x-bar < 504)

z = (484-494)/100/√100

z = -1 is 0.1587

z = (504-494)/100/√100

z = 1 is 0.8413

P(-1< z <1) = 0.6826

Yall are wrong

To solve these questions, we will use the concept of the standard normal distribution. The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1.

To find the probability, we will use the Z-score formula:
Z = (X - μ) / (σ / √n)
Where:
- X is the sample mean
- μ is the population mean
- σ is the population standard deviation
- n is the sample size

a. To find the probability for the Critical Reading part of the test:
μ = 502 (population mean)
σ = 100 (population standard deviation)
n = 90 (sample size)
We want to find the probability for a sample mean within 10 points of the population mean, so X = 502 ± 10.

Calculating the Z-scores:
Z1 = (502 - 502) / (100 / √90) = 0 (lower bound Z-score)
Z2 = (512 - 502) / (100 / √90) = 1 (upper bound Z-score)

Now we need to find the probabilities associated with these Z-scores using a standard normal distribution table or calculator. The probability is given by the area under the curve between the Z-scores - in this case, between 0 and 1.

Using a standard normal distribution table, we find that the probability is approximately 0.3413.

b. To find the probability for the Mathematics part of the test:
μ = 515 (population mean)
σ = 100 (population standard deviation)
n = 90 (sample size)
We want to find the probability for a sample mean within 10 points of the population mean, so X = 515 ± 10.

Calculating the Z-scores:
Z1 = (515 - 515) / (100 / √90) = 0 (lower bound Z-score)
Z2 = (525 - 515) / (100 / √90) = 1 (upper bound Z-score)

Using the same process as in part a, we find that the probability is approximately 0.3413.

c. To find the probability for the Writing part of the test:
μ = 494 (population mean)
σ = 100 (population standard deviation)
n = 100 (sample size)
We want to find the probability for a sample mean within 10 points of the population mean, so X = 494 ± 10.

Calculating the Z-scores:
Z1 = (494 - 494) / (100 / √100) = 0 (lower bound Z-score)
Z2 = (504 - 494) / (100 / √100) = 1 (upper bound Z-score)

Using the same process as in part a, we find that the probability is approximately 0.3413.

So, the probabilities are all approximately 0.3413 for parts a, b, and c.