Download Solutions to Test #2

Elementary Statistics
Fall, 2006
Name:
Dr. D. P. Story
Test 2
Instructions: (100 points) Solve each problem without error, feel free to use the probability tables
provided, or use your calculator to compute probabilities.
pts
(2ea. )
1. Multiple choice/fill-in the blank questions.
(a) Let x be the number of rainy days in the month of July in Florida. What type of random variable
is this? (Check one)
✔ discrete
continuous
(b) A binomial experiment has of fixed number of
independent
trials.
(c) The mean and standard deviation of the standard normal distribution are
respectively.
0
and
1
,
(d) If the size of a sample is at least
30
, you can use z-scores to determine the probability
that a sample mean falls in a given interval of the sampling distribution.
(e) As the size of a sample increases, the standard deviation of the distribution of the sample
means…
✔ decrease
increase
(5pts )
2. Determine the probability distribution’s missing probability value.
x
0
1
2
3
4
P (x)
0.07
0.20
0.38
0.22
0.13
Fill in the missing value.
Solution: The sum of the probabilities must be 1.04; thus,
Missing Value = 1 − (0.07 + 0.20 + 0.38 + 0.13) = 0.22
(7pts )
3. Compute the mean (expected value) of the random variable x, whose distribution is given below.
Label the additional row provided, and fill in the rest of the entries as part of the process of
computing the mean value of x.
μ=
1.67
x
0
1
2
3
Totals
P (x)
0.17
0.22
0.38
0.23
1.00
x · P (x)
0
0.22
0.76
0.69
1.67
Stats/T2
(6pts )
– Page 2 of 5 –
Name:
4. For the random variable and probability distribution given in Problem 3, compute each of the
probabilities given below:
P( 1 ≤ x ≤ 2 ) =
0.6
P ( x < 1 or x > 2 ) =
0.4
Solution: We add the probabilities given in the table in Problem 3:
P ( 1 ≤ x ≤ 2 ) = P (1) + P (2) = 0.22 + 0.38 = 0.6
P ( x < 1 or x > 2 ) = 1 − P ( 1 ≤ x ≤ 2 ) = 1 − 0.6 = 0.4
The second probability can be calculated directly as follows:
P ( x < 1 or x > 2 ) = P (0) + P (3) = 0.17 + 0.23 = 0.40
(5pts )
5. Let x be a binomial random variable on 15 trials with probability of success 0.35. Compute the
mean and variance of x.
μ=
σ2 =
5.25
3.4125
Solution: After noting that q = 0.65, we have…
μ = np = (15)(0.35) = 5.25
σ 2 = npq = (15)(0.35)(0.65) = 3.4125
(10pts )
6. Fifty-five percent of men consider themselves basketball fans. You randomly select 12 men and ask
each if he considers himself a basketball fan. Let x be the number of men who declared themselves
basketball fans. For parts (c) and (d), use either the binomial tables or your calculator.
Solution: Just look in the Binomial Tables, under n = 12, p = 0.55 and x = 7.
(a) (2 pts) n =
12
(b) (2 pts) p =
0.55
(c) (3 pts) Find the probability the number who
consider himself a basketball fan is exactly
seven.
Answer =
0.223
(d) (3 pts) P ( x > 9 ) =
0.043
Solution: From the Binomial Tables we have…
P ( x > 9 ) = P (10) + P (11) + P (12)
= 0.034 + 0.008 + 0.001
= 0.043
Stats/T2
pts
(3ea. )
– Page 3 of 5 –
Name:
7. Simple Problems for Standard Normal Distribution. Use either the standard normal tables or your
calculator to answer each of the following.
• In parts (a)–(c), find the indicated area under the standard normal curve.
(a) To the left of z = 0.25. Area =
0.5987
Solution: Direct table lookup, find z = 0.25 in the standard normal tables and read corresponding entry.
(b) To the right of −1.75. Area =
0.9599
Solution: Find z = −1.75 in tables, 0.0401. The answer is 1 − 0.0401 = 0.9599 .
(c) Between −1 and 1.5. Area =
0.7745
Solution: Look up the two z-values, 0.1587 and 0.9332, respectively, and subtract the smaller
from the larger to get the area between: 0.9332 − 0.1587 = 0.7745 .
• In parts (d)–(e), find the indicated probabilities.
(d) P ( z < 0.45 ) =
0.3264
Solution: Direct table look up, find the z-value for 0.45 to get an answer of 0.3264 .
(e) P ( z > −0.25 ) =
0.5987
Solution: Find z = −0.25 in tables, 0.4013. The answer is 1 − 0.4013 = 0.5987 .
pts
(4ea. )
8. In a recent year, the ACT scores for high school students were normally distributed with a mean
of 24.1 and a standard deviation of 4.3. A high school student who took the ACT during this time
is randomly selected.
(a) Find the probability that the student’s ACT score is less than 20.
Answer =
0.1711
Solution: Find the z-score first,
z=
20 − 24.1
≈ −0.95
4.3
Looking this z-value up in the standard normal tables we get 0.1711 .
(b) Find the probability that the student’s ACT score is between 20 and 29.
Answer =
0.7018
Solution: We begin by computing the z-score of each
20 − 24.1
29 − 24.1
≈ −0.95
z2 =
≈ 1.14
4.3
4.3
Using the standard normal tables, we look these two values up, 0.1711 and 0.8729. Subtracting
the small from the larger we get 0.8729 − 0.1711 = 0.7018 .
z1 =
Stats/T2
– Page 4 of 5 –
Name:
(c) What percentage of the high school population who took the ACT during this year have a score
of more than 29?
Answer =
12.71%
Solution: Compute the z-score of 29: z = (29 − 24.1)/4.3 ≈ 1.14, look up in standard normal
tables to get 0.8729. As we want the area to the right of this z-score, we subtract from 1 to
obtain 1 − 0.8729 = 0.1271, and finally converting to a percentage, 12.71% .
pts
(4ea. )
9. Find the z-score for each.
(a) Find the z-score that has 62.8 % of the distribution’s area to its right.
z-score =
−0.33
Solution: Reverse table look up. Find the number 1 − 0.628 = 0.372 in the body of the table,
and read off the corresponding z-score, which is about −0.33
(b) Find the z-score for which 12 % of the distribution’s area is between −z and z
z-score =
0.15
Solution: If −z and z enclose an area of 0.12, then there must be an area 0.44 in each of the
two tails (0.12 + 0.44 + 0.44 = 1). Therefore, the area to the left of z is 0.12 + 0.44 = 0.56. (A
picture would be useful.) Now, looking up 0.56 in the body of the standard normal tables, we
stumble across z = 0.15 (approximately). This is the answer z = 0.15 .
(4pts )
10. The annual per capita utilization of apples (in pounds) in the United States can be approximated
by the normal distribution with mean of 17.4 lb and standard deviation of 4 lb. What annual per
capita utilization of apples represents the 10th percentile?
P10 =
12.28
Solution: P10 has 10% of the area to the left of it. Looking up 0.10 in the body of the standard normal
tables and reading off the corresponding z-value we see z = −1.28. This is the 10th percentile of
the standard normal distribution. For the given distribution
P10 = μ + zσ = 17.4 + (−1.28)(4) = 12.28
(4pts )
11. In a survey of men in the United States (ages 20-29), the mean height was 69.2 inches with standard
deviation of 2.9 inches. What height represents the first quartile?
Q1 =
67.24
Solution: Recall Q1 = P25 . So solutions is the same as the previous problem. Q1 has 25% of the
area to the left of it. Looking up 0.25 in the body of the standard normal tables and reading off the
corresponding z-value we see z = −.675 (approximately). This is the Q1 of the standard normal
distribution. For the given distribution
Q1 = μ + zσ = 69.2 + (−.675)(2.9) ≈ 67.24
Stats/T2
(6pts )
– Page 5 of 5 –
Name:
12. The mean price of digital cameras at an electronics store is $349 with a standard devaition of $9.
Random samples of size 40 are drawn from this population. Compute the mean and standard
deviation of the sample mean.
μx̄ =
σx̄ =
349
1.42
Solution: We have
μx̄ = μ = 349
(8pts )
9
σ
≈ 1.42
σx̄ = √ = √
n
40
13. The mean height of women in the United States (ages 20-29) is μ = 64 inches. A random sample
of 60 women in this age group is selected. What is the probability that x̄, the mean height of the
sample, is greater than 64.5 inches? Assume σ = 2.75 inches.
P ( x̄ > 64.5 ) =
0.08
Solution: First compute the mean and standard deviation of x̄:
μx̄ = μ = 64
2.75
σ
≈ 0.355
σx̄ = √ = √
n
60
Now, using these values, compute the z-value of 64.5:
64.5 − 64
≈ 1.41
0.355
Looking this z-value up in the standard normal tables, we get 0.9207. Finally, as we want the area
to the right, we subtract from 1 to get 1 − 0.9207 ≈ 0.08
z=