Statistics in Business
Sections 4 and 5
Class 05 Assignment
Due Wed February 1, 2012
NAME:_________________________
1. (EMBS problem 19, page 260) The average amount of precipitation in Dallas, Texas during April is 3.5
inches (The World Almanac, 2000). Assume that a normal distribution applies and that the standard
deviation is 0.8 inches.
a. What percentage of Aprils do we expect precipitation to exceed 5 inches?
b. What percentage of Aprils do we expect less than 3 inches?
c. A month is classified as extremely wet if the amount of precipitation is in the upper 10% for that
month. How much precipitation must there be in April for it to be classified as extremely wet?
a. = 1-normdist(5,3.5,.8,true) = 0.030 = 3.0%
b = normdist(3,3.5,.8,true) = 0.266 = 26.6%
c = norminv(.9,3.5,.8) = 4.525 inches
2. (EMBS problem 23, page 260) The time needed to complete a final exam in a particular college course
is normally distributed with a mean of 80 minutes and a standard deviation of 10 minutes.
a. What is the probability a student completes the exam in one hour or less?
b. What is the probability the student will complete the exam in more than 60 but less than 75 minutes?
c. The distribution of number of hours to complete this exam is normal with a mean of 1.333. What is
the variance?
d. Assume the class has 60 students and the period is only 90 minutes long. What percentage of
students do we expect will not be able to finish the exam?
e. (extra credit) How many of the 60 will not be able to complete the exam in the 90 minutes? (State any
additional assumptions you make in answering part e.)
a. = normdist(60,80,10,true) = 0.023
b = normdist(75,80,10,true) – normdist(60,80,10,true) = 0.309 – 0.023 = 0.286
c = the standard deviation in hours is 10/60. The variance in hours is the square of this = 0.028
d.=1-normdist(90,80,10,true) = 0.159
e. The number not finishing is a random variable. If completion times are independent and identically
distributed across the 60 students, then the number not finishing will be binomially distribution with
n=60 and p = 0.159
3. Cy is the firm’s expert forecaster who uses the language of probability when making forecasts. With
respect to next year’s total sales, Cy’s best guess is $4.5 million, and Cy says there is a 95% probability
total sales will be between $3.5 and $5.5 million. If the normal distribution applies, what is the
probability sales will exceed $5 million next year?
If the normal distribution applies, then the best guess of $4.5 is the mean. To find the standard
deviation, we use the fact that there is about a 95% probability the random variable will be within +/two (actually it is 1.96) of the mean. This translates to a standard deviation of ($5.5 - $4.5)/2 = $0.5
million. The probability sales will exceed $5 is then 1 – normdist(5,4.5,.5,true) = 0.159.
4. The filling machine is set to fill bottles with a target (mean) amount of 10.2 fluid ounces. The actual
amount in a bottle will be normally distributed with mean 10.2 ounces and standard deviation of 0.16
ounces. If the machine fills 100 bottles, how many bottles do we expect in each of the following seven
bins? In other words, how many of the 100 bottles do we expect will contain less than 9.95 ounces?
How many do we expect will contain more than 9.95 and less than 10.05? And so on…..
BIN
Expected Count
< 9.95
5.91
9.95 to 10.05
11.52
10.05 to 10.15
20.31
10.15 to 10.25
24.53
10.25 to 10.35
20.31
10.35 to 10.45
11.52
> 10.45
5.91
The expected count in any cell will be 100*[NORMDIST(upperbinlbound,10.2,0.16,true)normdist(lowerbinbound,10.2,.16,true)]. So for the second bin, we have =
100*[normdist(10.05,10.2,0.16,true)-normidst(9.95,10.2,0.16,true)] = 100*(0.174-0.059) = 11.52
bottles.