Chapter 6, Q18

A stock exchange has a mean stock price of $20 with standard deviation of $8.20.

Q. What is the probability a company will have a stock price of at least $40?

A. We want the right side of the distribution, because of the words at least.
So = 1 – norm.dist(40,20,8.20, true)= 0.007363

Q. How high does the stock price have to be to put the company in the top 10%?

A. The top 10% means on the right side of the distribution.

We need to find the value of the random variable that separates the top 10% from the bottom 90%.

We can use norm.inv for this. But—careful---if you put =norm.inv(0.1,20,8.2) you will get the random variable that separates the bottom 10 % from the top 90 %.

So =norm.inv(0.9,20,8.2) = 30.51 (rounded).

Chapter 5 Q32 Military Detection

Chapter 5 Q 32

Probability of one system detecting an attack is 0.9.

a. Probability of one system detecting an attack? 0.9 of course.

b. Probability of two systems installed in the same area and operating independently detecting an attack?

You could solve this using a tree diagram, but using binomials it like this: we have two trials because there are two systems, identical and independent. We want at least one success. That means we don’t want zero successes. So find the probability of zero successes and then subtract from 1. That will give us "at least one".

=binomdist(0,2,0.9, false) = 0.01

Then 1 – 0.01 = 0.99

c. Three systems is the same argument

=binomdist(0,3,0.99,false) = 0.001

Then 1 – 0.001 = 0.999

d. Would you recommend that multiple systems be used?

It would depend on the cost of extra systems and the likelihood of an attack in the first place. If the system is expensive and the risk of an attack is low, then the gain from 0.99 to 0.999 is very small compared to say, spending that money on better education for kids.

Testing for independence: probability

Q. If P(A) = 0.5, P(B) = 0.6 and P(A∩B)=0.3, test whether A and B are independent.

A. The test for independence is: if P(A|B) = P(A) then A and B are independent.

To run the test, we need to find P(A|B). We get this from P(A|B) = P(A∩B)/P(B).

Here that number is  0.3/0.6 = 0.5.

Now, is the correct? Does P(A|B) = P(A)? Yes because 0.5 = 0.5. So A and B are independent. 

Chapter 6 Q34

Chapter 6, Q 34  page 279. There is a Youtube here that might help


Question: Family annual expenditure on food and drink( normally distributed), mean=5700, standard deviation=1500


a. What is the range of expenditures of the 10% of the families with the lowest annual expenditure on food and drink?


The 10% is the lowest, so that is on the left side of the distribution. Use =norm.inv(0.1,5700,1500) to give 3777.67. So expenditure would be 3780 (rounding) or less.


b. What percentage of family spend more than $7000 annually on food and drink?


More than means expenditure is greater than $7000. Recall that Excel adds up from the left. So, we will need to subtract from 1. So go =1-norm.dist(7000,5700,1500,true) which gives 0.19. There is a blog posting here

 c. What is the range of expenditures for the 5% of families with the highest annual spending on food and drink?


The highest annual expenditure must be on the right of the distribution. It is the little ‘tail’. Recall that Excel adds up from the left. So if you use norm.inv with 0.05, you will get the bottom 5%. So use 0.95, and that will give you the value of the random variable (here annual expenditure) that separates the bottom 95% to the top 5%. Blog posting
=norm.inv(0.95,5700,1500) which gives $8167.28

Chapter 7 Q22

Q. The annual cost of automobile insurance is $939. Assume that the population standard deviation is $245. Find the probability that a SRS of insurance policies will have a sample mean within $25 of the population mean for sample sizes 30, 50, 100, and 400.

A. The interval is between 939 – 25, and 939 + 25 or 914 to 964. We can do this in Excel. For the first sample, where n =30, find the standard error. This is the population standard deviation divided by the square root of the sample size. Where n =30, the standard error is 44.73.

First find the probability from the extreme left of the distribution to 964. In Excel that is =norm.dist(964,939,44.73,true) = 0.71. 

Now find the probability from the extreme left of the distribution to 914. In Excel this is =norm.dist(914,939,44.73,true) = 0.29. 

The final step is to subtract the smaller probability from the larger one. This gives the probability of the area between 914 and 964. This is 0.71 – 0.29 = 0.42.

Follow the same steps with the larger samples. You will see that the probability increases with the sample size. As the sample size increases, the standard error decreases and we are more confident of the location of the unknown population parameter µ. For example, with a sample size of n=400, the probability increases to 0.96.

In this question we know µ. But make the intellectual leap to see that we can use the same method to estimate the location of µ if we did not know it. 

Chapter 6 Q20

An alert student has noticed a mistake in the textbook answers. Here I've worked through Q20 from Chapter 6.

a.       a. We want the area to the left of 50, because the question asks for fewer than 50 hours. Use =norm.dist(50,77,20,true) to get 0.089

b.      b. We want more than 100, the area to the right of 100. Excel adds up from the left, so we need to subtract from 1. Use =1-norm.dist(100,77,20,true) to get 0.125

c.       c. The question asks for the “upper 20%”. We are asked to find a random variable which corresponds to this area. In other words, what is the value of the random variable X which separates the top 20% from the bottom 80%. We can use =norm.inv for this, but we will need to write =norm.inv(0.8,77,20) to get 93.83. If we had written =norm.inv(0.2,77,20) then the result would the value of the random variable X that separates the bottom 20% from the top 80%. Draw a sketch and make sure you get this. This posting on our blog might help make this clearer.

Q4 p333

Q4 p333

Q: A 95% confidence interval for a population mean was reported to be 152 to 160. If σ = 15, what sample size was used in this study?

A: The mean must be right in the middle of the CI, so here the mean, xbar, is 156. Therefore the Margin of Error was 4. Recall how we find the MofE in the first place? It is MofE =1.96*sigma/root n.

So in this case 4=(1.96 *15)/root n.

Switch the terms around so that root n = (1.96*15)/4 = 7.35 That’s the square root of the sample size, so square it to get n =54.02. ROUND UP to get n = 55