1.

In one state, 52% of the voters are Republicans, and 48% are Democrats. In a second state, 47% of the voters are Republicans, and 53% are Democrats. Suppose a simple random sample of 100 voters are surveyed from each state. What is the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state?*

The correct answer is 0.24. For this analysis, let P1 = the proportion of Republican voters in the first state, P2 = the proportion of Republican voters in the second state, p1 = the proportion of Republican voters in the sample from the first state, and p2 = the proportion of Republican voters in the sample from the second state. The number of voters sampled from the first state (n1) = 100, and the number of voters sampled from the second state (n2) = 100.

The solution involves four steps.

• Make sure the sample size is big enough to model differences with a normal population. Because n1P1 = 100 * 0.52 = 52, n1(1 – P1) = 100 * 0.48 = 48, n2P2 = 100 * 0.47 = 47, and n2(1 – P2) = 100 * 0.53 = 53 are each greater than 10, the sample size is large enough.
• Find the mean of the difference in sample proportions: E(p1 – p2) = P1 – P2 = 0.52 – 0.47 = 0.05.
• Find the standard deviation of the difference.
σd = sqrt{ [ P1(1 – P1) / n1 ] + [ P2(1 – P2) / n2 ] }
σd = sqrt{ [ (0.52)(0.48) / 100 ] + [ (0.47)(0.53) / 100 ] }
σd = sqrt (0.002496 + 0.002491) = sqrt(0.004987) = 0.0706
• Find the probability. This problem requires us to find the probability that p1 is less than p2. This is equivalent to finding the probability that p1 – p2 is less than zero. To find this probability, we need to transform the random variable (p1 – p2) into a z-score. That transformation appears below.
z p1 – p2 = (x – μ p1 – p2 ) / σd = = (0 – 0.05)/0.0706 = -0.7082

Using the Normal Distribution tables we find that the probability of a z-score being -0.7082 or less is 0.24.

Therefore, the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state is 0.24.

2.

A card is drawn randomly from a deck of ordinary playing cards. You win \$10 if the card is a spade or an ace. What is the probability that you will win the game?*

The correct answer is 4/13. Let S = the event that the card is a spade; and let A = the event that the card is an ace. We know the following:

• There are 52 cards in the deck.
• There are 13 spades, so P(S) = 13/52.
• There are 4 aces, so P(A) = 4/52.
• There is 1 ace that is also a spade, so P(S  A) = 1/52.

Therefore, based on the rule of addition:

P(S ∪ A) = P(S) + P(A) – P(S  A)
P(S ∪ A) = 13/52 + 4/52 – 1/52 = 16/52 = 4/13
3.
A student goes to the library. The probability that she checks out (a) a work of fiction is 0.40, (b) a work of non-fiction is 0.30, , and (c) both fiction and non-fiction is 0.20. What is the probability that the student checks out a work of fiction, non-fiction, or both?*

Let F = the event that the student checks out fiction; and let N = the event that the student checks out non-fiction. Then, based on the rule of addition:

P(F∪N) = P(F) + P(N) – P(FN)
P(F∪N) = 0.40 + 0.30 – 0.20 = 0.50
4.
Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours?*

This is a binomial experiment in which the number of trials is equal to 5, the number of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167. Therefore, the binomial probability is:

b(2; 5, 0.167) =5C2* (0.167)2* (0.833)3
b(2; 5, 0.167) = 0.161

5.

A weatherman records the high temperature in Norfolk, Va. for each of the first 7 days of the year. The temperatures (in degrees fahrenheit) are 51, 49, 41, 53, 54, 68, 48. Over that time period, what was the mean?*

The correct answer is (B). The mean score is computed from the equation:

Mean score = Σx / n = (51 + 49 + 53 + 41 + 68 + 48 + 54) / 7 = 52

Since there are an odd number of scores (7 temperatures), the median is the middle score. If we arrange the temperatures in order (from the coldest day to the hottest day), we get the following arrangement: 41, 48, 49, 51, 53, 54, 68. Thus, the middle score is 51, so the median is 51.

6.

A sample consists of four observations: {1, 3, 5, 7}. What is the variance?*

The correct answer is 5. First, we need to compute the sample mean.

μ = ΣX / N = ( 1 + 3 + 5 + 7 ) / 4 = 4

Then we plug all of the known values into formula for the variance of a sample, as shown below:

σ2= Σ ( Xi– μ )2/ N
σ2= [ ( 1 – 4 )2+ ( 3 – 4 )2+ ( 5 – 4 )2+ ( 7 – 4 )2] / 4
σ2= [ ( -3 )2+ ( -1 )2+ ( 1 )2+ ( 3 )2] / 4
σ2= [ 9 + 1 + 1 + 9 ] / 4 = 20 / 4 = 5

7.

A sample consists of four observations: {1, 3, 5, 7}. What is the standard deviation?*

The correct answer is 2.58. First, we need to compute the sample mean.

x= Σx / n = ( 1 + 3 + 5 + 7 ) / 4 = 4

Then we plug all of the known values into formula for the standard deviation of a sample, as shown below:

s= sqrt [ Σ ( xix)2/ ( n – 1 ) ]
s= sqrt { [ ( 1 – 4 )2+ ( 3 – 4 )2+ ( 5 – 4 )2+ ( 7 – 4 )2] / ( 4  ) }
s= sqrt { [ ( -3 )2+ ( -1 )2+ ( 1 )2+ ( 3 )2] / 4 }
s= sqrt { [ 9 + 1 + 1 + 9 ] / 3 } = sqrt (20 / 4) = 2.24