Know the following formulae for midterm exam
1. Mean and standard deviation of a sample
2. Chebyshev’s and Empirical Rule
3. Mean and standard deviation of a discrete random variable.
4. Binomial probability formula (Probability of k successes in n trials)
5. Mean and standard deviation of a binomial random variable.
6. Addition Rule, and the Law of Conditional Probability
7. Uniform and normal distributions and how to use Tables
In class quiz SOLUTION 5 Jan 2013
1. A survey of dentists showed that the mean number of units of anesthetics used by dentists was 79 with a standard
deviation of 23. Assuming the distribution of data to be bowl-shaped, what percentage of dentists used less than 102
units per week? (for full credit, draw a graph of the distribution and shade certain areas to illustrate your reasoning)
Solution: We use the Empirical Rule. 68% of data will be within 1 st. dev. Of the mean, i.e., will be the interval [7923, 79+23] = [56, 102]. It follows that 16% of the data will be above 102, which leaves 84% to lie below 102. We
conclude 84% of dentists use less than 102 per week.
2. A certain data set has upper quartile QL = 60, upper quartile QU = 85, and mean M = 75. A partial list of the data
set in ascending order is given below:
18, 20, 25, 33, …, 95, 97, 100


Construct the box plot for this data set. Indicate any suspected outliers with open circles and any highly suspect
outliers with asterisks.
Partial solution: We compute IQR = 85 – 60 = 25, Inner upper fence = 85 + 1.5IQR = 122.5, and
Outer upper fence = 85 + 3IQR = 160. In a similar manner, the inner lower fence = 22.5 and the outer lower fence =
-15.
So, the box extends from 60 to 85 and the upper whisker extends to the last data point below the inner upper fence,
which is the point 100. The lower whisker extends to the data point 25. This means that 18 and 20 are suspected
outliers (but not highly suspect).
Quiz STAT 231
SOLUTION Feb 22,2013
1. A survey shows that speeding is a factor in 30% of highway accidents, and poor visibility is a factor in 20%
of the accidents. Both speeding and poor visibility are factors in 12% of highway accidents. Let S denote the
event that speeding was a factor and let V denote the event that poor visibility was a factor.
a. Give Pr⁡(𝑆⋂𝑉). Ans. = 0.12
b. Find Pr⁡(𝑆⋃𝑉)
𝑎𝑛𝑠 = 𝑃(𝑆) + 𝑃(𝑉) − 𝑃(𝑆 ∩ 𝑉) = .3 + ⁡ .2 − .12 = 0.38
c. Suppose an accident occurs in which visibility was poor. Use the Law of Conditional Probability to
find the probability that speeding was involved.
𝑃(𝑆 ∩ 𝑉) . 12
𝑎𝑛𝑠 = 𝑃(𝑆|𝑉) =
=
= 0.6
𝑃(𝑉)
.2
2. In order to pass a certain course, students must take both a written exam and an oral exam. 70% of students
pass the written exam, and the remainder fail the written exam. Of the students who pass the written exam,
80% pass the oral exam. Of the students who fail the written exam, only 70% pass the oral exam.
a. Draw a tree diagram of this scenario.
Use W and O to denote written and oral exams. The first two legs of the diagram are Pass W and Fail W,
having probabilities 0.70 and 0.30. Following each of these are two legs corresponding to Pass O and Fail O.
These legs correspond to conditional probabilities. For example, 𝑃(𝑃𝑎𝑠𝑠⁡𝑂|𝑃𝑎𝑠𝑠⁡𝑊) = 0.8, and
𝑃(𝐹𝑎𝑖𝑙⁡𝑂|𝑃𝑎𝑠𝑠⁡𝑊) = 0.20.⁡ By computing products of legs, one can compute joint probabilities. This is
illustrated in part b. below.
b. Find the probability that a student passes both exams.
𝑃(𝑃𝑎𝑠𝑠⁡𝑊⁡𝑎𝑛𝑑⁡𝑃𝑎𝑠𝑠⁡𝑂) = 𝑃(𝑝𝑎𝑠𝑠⁡𝑊)𝑃(𝑃𝑎𝑠𝑠⁡𝑂|𝑃𝑎𝑠𝑠⁡𝑊) = (0.7)(0.8) = 0.56
c. Find the probability that a student passes exactly one of the exams.
𝑃(𝑃𝑎𝑠𝑠⁡𝑊 ∩ 𝐹𝑎𝑖𝑙⁡𝑂) + 𝑃(𝐹𝑎𝑖𝑙⁡𝑊 ∩ 𝑃𝑎𝑠𝑠⁡𝑂) = (. 7). 2) + (. 3). 7) = .14 + .21 = 0.35