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