Question

An article in the October 11, 2006, issue of the Washington Post claimed that 15% of high school students used cursive writing on the essay portion of the SAT exam in the academic year 2005-2006 (Pressler, 2006). Suppose you take a random sample from those exams and see what proportion of the sample used cursive writing for the essay. Assume the sample size is 180 do the following:

• A -- Determine what π and N are.
• B -- Describe the sampling distribution.
• C -- Show that the CLT holds.
• D -- What does the CLT say about the sampling distribution?
• E -- What is the probability of obtaining a sample where more than half of the students wrote their essay in cursive?
• F -- What is the probability of obtaining a sample where between 10 and 25% wrote their essay in cursive?
• G -- What is the probability of obtaining a sample where less than 12% of the students wrote their essay in cursive?

Answer 1

Solution:

Given:

Claim: 15% of high school students used cursive writing on the essay portion of the SAT examination in the academic year 2005-2006

Sample Size = 180

Part A) Determine what π and N are.

proportion of high school students used cursive writing on the essay portion of the SAT examination in the academic year 2005-2006

and

N = Sample Size = 180

Part B) Describe the sampling distribution.

Since sample size N = 180 is large, so sampling distribution of sample proportion would be approximately Normal.

Part C) Show that the CLT holds.

CLT means Central limit theorem.

Assumptions of CLT are:

and

Thus we get:

and

Since both assumptions are satisfied, CLT holds.

Part D) What does the CLT say about the sampling distribution?

CLT shows that the sampling distribution of sample proportion is approximately Normal with mean of sample proportions is and standard deviation of sample proportion is :

Part E) What is the probability of obtaining a sample where more than half of the students wrote their essay in cursive?

That is we have to find:

Thus find z score :

Thus we get:

Since Z = 13.15 is too large and area under z =13.15 is approximately 100% that is: 1.0000

That is: P(Z < 13.15) = 1.0000

Thus we get:.

Thus the probability of obtaining a sample where more than half of the students wrote their essay in cursive is 0.0000

Part F) What is the probability of obtaining a sample where between 10% and 25% wrote their essay in cursive?

That is we have to find:

Find z scores for 0.10 and 0.25:

Thus we get:

Look in z table for z = 3.7 and 0.06 as well as for z = -1.8 and 0.08

and find corresponding area.

P( Z < 3.76) = 0.99992

P( Z < -1.88) =0.0301

Thus we get:

Part G) What is the probability of obtaining a sample where less than 12% of the students wrote their essay in cursive?

Thus we get:

Look in z table for z = -1.1 and 0.03 and find area.

P( Z < -1.13) = 0.1292

Thus we get: