MM207 Unit 4 Sample Problems
Similar Examples:
1. Decide whether the random variable is discrete or continuous.
(References: example 1 page 195, end of section exercises 13 – 20 page 201)
a.
The amount of waste a sewage treatment plant handles monthly
Answer: continuous (has an uncountable number of possible outcomes, represented by an
interval on the number line)
b.
The weight of a hummingbird
Answer: continuous (has an uncountable number of possible outcomes, represented by an
interval on the number line)
c.
The resting heart rate of an athlete
Answer: discrete (has a finite or countable number of possible outcomes that can be listed)
d.
The number of winners in a dance contest
Answer: discrete (has a finite or countable number of possible outcomes that can be listed)
e.
The length of time between births at the Mayo Clinic on January 1, 2009
Answer: continuous (has an uncountable number of possible outcomes, represented by an
interval on the number line)
Similar Example:
2. Decide whether the distribution is a probability distribution. If it is not a probability distribution,
identify the property that is not satisfied. (References: example 3 and 4 page 197, end of
section exercises 25 - 28 page 202 - 203)
x
1
2
3
4
5
P(x)
0.165
0.189
0.211
0.287
0.148
Solution:
According to the definition on p. 195 of the text, a probability distribution must satisfy the
following conditions:
i.
The probability of each value of the discrete random variable is between 0 and 1, inclusive
0 ≤ P(x) ≤ 1, and
ii.
The sum of all the probabilities is 1.
 P(x) = 1
This distribution satisfies both of these requirements, so it is a probability distribution.
Similar Example:
3.
The table below shows the probability distribution for the number of dogs per household in a
suburban area. Complete the table shown and find the mean, variance, and standard deviation
for the distribution. Round b, c and d to 2 decimal places as needed.
(References: example 5 and 6 page 198 - 199, end of section exercises 29 - 34 page
203)
1
MM207 Unit 4 Sample Problems
x
P(x)
0
1
2
3
4
0.54
0.23
0.12
0.07
0.04
x  P(x)
x-
(x - )2
P(x)  (x - )2
Round to 2
decimals
Round to 2
decimals
Round to 4
decimals
Round to 4 decimals
0.00
0.23
0.24
0.21
0.16
 x  P(x) =
0.84
-0.84
0.16
1.16
2.16
3.16
0.7056
0.0256
1.3456
4.6656
9.9856
0.3810
0.0059
0.1615
0.3266
0.3994
 P(x)  (x - )2 =
1.2744
a. Complete the Table
b. Mean 
Solution: Mean  =  x  P(x) = 0.84 (total from table)
c. Variance
2
Solution: Variance
2 =  P(x)  (x - )2 = 1.2744  1.27 (total from table)
d. Standard Deviation σ
Solution: σ = √1.27 = 1.13
Similar Example:
4.
At one soft drink distributing plant, 45% of the aluminum cans used are produced out of state.
20 cans are chosen at random from a shipment headed to a local grocery store.
a. What is the probability that exactly 4 cans were produced out of state?
Solution: Refer to Table 2 – Binomial Distribution in the text.
For n = 20, p = 0.45, P(x = 4) = 0.014
Or, using Excel, BINOMDIST(4,20,0.45,FALSE)=0.014
b.
What is the probability that 4 or fewer can were produced out of state?
Solution: Refer to Table 2 – Binomial Distribution in the text.
For n = 20, p = 0.45, P(x  4) = P(0) + P(1) + P(2) + P(3) + P(4)
= 0.000 + 0.000 + 0.001 + 0.004 + 0.014 = 0.019
Or, using Excel, BINOMDIST(4,20,0.45,TRUE)=0.019
c.
What is the probability that exactly 10 can were produced out of state?
Solution: Refer to Table 2 – Binomial Distribution in the text.
For n = 20, p = 0.45, P(x = 10) = 0.159
Or, using Excel, BINOMDIST(10,20,0.45,FALSE)=0.159
d.
Find the mean 
Solution:  = np = 20(0.45) = 9
e.
Find the variance
Solution:
2
2
 = npq = 20(0.45) (0.55) = 4.95
Similar Example:
5.
A true-false test has 10 questions. Assume a student randomly guesses the answer to every
question.
a. What is the probability of getting exactly 5 correct answers?
Solution: Refer to Table 2 – Binomial Distribution in the text.
For n = 10, p = 0.50, P(x = 5) = 0.246
Or, using Excel, BINOMDIST(5,10,0.50,FALSE)=0.246
b. What is the probability of getting fewer than 7 correct answers?
2
MM207 Unit 4 Sample Problems
Solution: For n = 10, p = 0.50,
P(x < 7) = P(x  6) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
= 0.001 + 0.010 + 0.044 + 0.117 + 0.205 + 0.246 + 0.205 = 0.828
Or, using Excel, BINOMDIST(6,10,0.50,TRUE)=0.828
3