3. There were 200 people who took the SAT examination; given their SAT verbal sc

Question

3. There were 200 people who took the SAT examination; given their SAT verbal score in which the mean is 500 and the standard deviation is 100, compute the following:

a. What percentage of people would an individual beat with a score of 600?

b. What percentage of people would an individual beat with a score of 370?

c. What percentage of people fall between 420 and 560?

d. What score would 4% of the people be above?

4. For a sample with a mean of 50 and a standard deviation of 5, what is the 95% confidence interval on the mean?

5. What is the 99% confidence interval on the mean for the same sample?

6. In a 2009 Nielsen survey, it was stated that Americans (2 years and older) watch television on average 35 hours per week with a standard deviation of 4.2. It seems likely that high school and college students don't watch nearly this much television per week. You survey a sample of 15 high school and college students and find out that they watch an average of 20 hours of TV per week.

a. What is the z-score for the cut-off region if we were to reject the null hypothesis at 5%?

b. Compute the z-score for the sample’s mean.

c. Do you reject or fail to reject the null hypothesis?

d. What would be your conclusion about the amount of TV that high school and college students watch per week?

Explanation / Answer

Solution:

We are given n=200

Mean = 500

SD = 100

Part a

We have to compute the percentage of people who beat a score of 600.

That is, we have to find P(X>600)

P(X>600) = 1 – P(X<600)

Z = (X – mean) / SD

Z = (600 – 500) / 100

Z = 100/100

Z = 1

P(Z<1) = P(X<600) = 0.841345(By using normal table)

P(X>600) = 1 – P(X<600)

P(X>600) = 1 – 0.841345

P(X>600) = 0.158655

Required percentage = 15.87%

Part b

We have to compute the percentage of people who beat a score of 370.

That is, we have to find P(X>370)

P(X>370) = 1 – P(X<370)

Z = (X – mean) / SD

Z = (370 – 500) / 100

Z = -130/100

Z = -1.3

P(Z<-1.3) = P(X<370) = 0.0968 (By using normal table)

P(X>370) = 1 – P(X<370)

P(X>370) = 1 – 0.0968

P(X>370) = 0.9032

Required percentage = 90.32%

Part c

Here, we have to find P(420<X<560)

P(420<X<560) = P(X<560) – P(X<420)

First we have to find P(X<560)

Z = (X – mean) / SD

Z = (560 – 500) / 100

Z = 60/100

Z =0.6

P(Z<0.6) = P(X<560) = 0.725747

Now, we have to find P(X<420)

Z = (420 – 500) / 100

Z = -80/100

Z = -0.8

P(Z<-0.8) = P(X<420) = 0.211855 (By using normal table)

P(420<X<560) = P(X<560) – P(X<420)

P(420<X<560) = 0.725747 - 0.211855

P(420<X<560) = 0.513892

Required percentage = 51.39%

Part d

Here, we have to find X-score for above 4% of the people or below 96% of the people.

X = Mean + Z*SD

Z for above 4% or below 96% is 1.750686 (By using normal table).

X = 500 + 1.750686*100

X = 675.0686

Required score = 675.07 approximately

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