Spring 2012
Practice Problems – Chapters 6, 7, 8
1. Find the mean and standard deviation for the following probability distribution:
x
1
2
3
P(x)
0.12
0.15
0.73
2. A raffle offers one $1000 prize, one $500 prize, and one $100 prize. One thousand
tickets are sold. You purchase one ticket for $3. Calculate your expected winnings.
Outcome
Winnings
Probability
3. For a sample of 30 teenagers, 70% of them had MP3 players by the age of 16.
Find the following probabilities:
a) Exactly 18 had an MP3 player by the age of 16.
b) Less than 17 had an MP3 player by the age of 16.
c) At least 14 had an MP3 player by the age of 16.
4. In a sample of 450 bananas, the probability of a banana being rotten is 6%.
a) Calculate the mean and standard deviation for the number of rotten
bananas in this sample.
b) Suppose that a produce manager found 35 rotten bananas in the sample of 450.
Is this usual or unusual and why?
5. Find the value of A if P(z > A) = .1056.
6. The salaries for full-time PhD students are normally distributed with an average of $12,837
per year and a standard deviation of $1500.
a) Find the probability that a randomly selected student will make more than $15,000.
b) If a group of 50 graduate students are selected, find the probability that the average
salary is more than $15,000.
7. To qualify for a job at a health club, candidates must score in the upper 40% on a physical
fitness test. The test has a mean score of 200 and a standard deviation of 20. Find the
lowest possible qualifying score. Round your answer to the nearest whole number.
ANSWERS:
1. Use 1-VarStats(L1, L2)
2. Use 1-VarStats(L1, L2)
Mean = 2.61, St.Dev. = .69
E(x) = –$1.39
3. a) binompdf (30, .70, 18) = 0.0749
b) binomcdf (30, .70, 16) = 0.0401
c) 1 – binomcdf (30, .70, 13) = 0.9979
Outcome
Winnings
Probability
Win $1000
1000
1/1000
Win $500
500
1/1000
Win $100
100
1/1000
4. n = 450, p = 0.06, q = 0.94
Lose $3
–3
997/1000
a) Mean = 27, St.Dev = 5
b) 35 is not unusual because it lies within 2 standard deviations of the mean.
<------ µ–2σ-----µ–σ------µ------µ+σ------µ+2σ--------->
17
22
27
32
37
5. The value of A must be on the positive side of the graph since 0.1056 lies to the right of it.
The area to the left of A is 1 – 0.1056 = .8944. invNorm(.8944) = 1.25 A = 1.25
6. a) P(x > 15000) = normalcdf (15000, E99, 12837, 1500) = .0747
b) P( x > 15000) = normalcdf (15000, E99, 12837, 1500
50
) = .0000 +
7. “Upper 40%” means that 40% is above the cutoff and 60% is below it.
Note that the cutoff will be on the right side of the graph (above the mean).
invNorm(.60, 200, 20) = 205