Exam # 3 / Problem # 7

Mohammed Ahmed ( Prof I spoke to you in class on Thursday regarding the mistake on question # 25 , you told me I have the weekend to to post a new question, thank you)

Step 1: The first step in solving this problem is using the  right formula, the correct formula for this problem would be x-u divided by o. X being the given , U being the average and O being the standard deviation. My iPad does not have the actual symbols but the correct mathematical symbol for U and O has a line at he corner.

Step 2: We take our given numbers from our problem and simple just plug them in.  100-134/20= -1.7 and 150-134/20= 0.8

Step 3: Lastly we find the z score of both results and subtract them and get our final answer.

(Z score of 0.8) .7881- .0446 ( Z score of -1.7) = .7435

Study Guide for Problem #7 on Exam #3 Review Sheet

I’m Laticia Bourne, and I’ll be giving a step by step explanation on how to solve problem #7 on Exam #3’s review sheet, which reads:

The quarterly profits for a large company are normally distributed with  million and million.  What is the probability that the profits for the next quarter will lie between $100 million and $150 million?

Step 1.

The first thing that’s always best to do is pull out the information given.

Mean: 134 million

Standard Deviation: 20 million

In this case, we have two x’s

X = 150

X= 100


Step 2.

Before we can find the probability, we first have to find two z-scores for our two x’s. We do this by using the following equation:

z = x – mean/standard deviation

We’ve already pulled out the information needed for that equation in Step 1, so now to plug those numbers in for each of our x’s:

1. z = 150-134/20

z= 16/20

z = 0.8

2. z = 100-134/20

z= -34/20

z = -1.7


Step 3.

Now that we have our two z-scores, we then use the z-table to look up the area for each one. You should find that the area for the z-score 0.8 is .7881 and the area for the z-score -1.7 is .0446.


Step 4.

The final step to finding the probability to our problem is to now subtract our areas:


So the answer to our problem is:

The probability that the profits for the next quarter will lie between $100 million and $150 million is .7435.


I hope my explanation was very helpful, good luck on the final everyone! 🙂

Study Guide for Problem #14

Candice Wright

14. One tire manufacturer claims that his tires last an average of 42,000 miles with a standard deviation of 7800 miles. A random sample of 100 of his tires is taken. What is the probability that the average of these 100 tires will last greater than (to the right of) 41,000 miles?

First: We figure out what type of problem the question is and what is the question asking.

This is a sampling means distribution question (the same as day 20’s handout.) We know this because it is stated in the question, underlined above (random sample). The question is asking for the probability of 100 tires greater than 41,000 miles, so this means that our answer has to be between 0 and 1.

Second: We find the mean(µ), n, x-bar, and standard deviation S/σ where the mean/average (µ) stated above is 42000, n is 100, S/σ also stated above is 7800 once we know these numbers we could plug them into the formulas below to solve the problem.   (formulas used included)

Mean (µ) = 42,000

S/σ= 7,800

n = 100

x-bar = 41,000

Formulas Needed 

1. Sampling Distributions: for standard deviation σx-bar: S/√n

2. Sampling Distributions  convert x-bar to z:  z = (x-bar – μ) / σx-bar

*We use the sampling distribution formulas rather than the normal distribution formulas because we are working with samples.

 Third: Construct the Bell curve and map out which side of the curve we are finding the probability for. (see inserted picture). We have to find the probability to the RIGHT of the bell curve so we have to subtract 1 (one) from the x’s Z score. Why? Because the table only gives us the Z score to the left not the right.

Solving the problem

Step 1: We find the standard deviation for x-bar first because this answer is used as the denominator for the following step.

σx-bar = S/√n=  7800/√100 = 7800/10 = 780

Step 2: we use the sampling distribution formula for x-bar to convert our x to z.

z = (x-bar – μ) / σx-bar = 41000-42000/780 =(41000-42000=-1000) -1000/780=  -1.28 (round answer to nearest two decimals)

*Note: we round the answer to the nearest two decimal places because we are finding the Z score and we only need two decimal places.

Step 3: We look up the Z score in the table

-1.28 –> .1003

 Step 4: .1003 would have been our answer if we were finding x to the left of the bell curve. In this case we have an extra step and that is to find x to the right of the bell curve. There are two ways in finding the Z score to the right of a bell curve.

1. Take the Z score and switch the sign. (it will either be positive of negative so the opposite sign of whatever your Z score is)

-1.28 –>1.28

Then we will just look up the Z score in the table and find our answer (x).

1.28 –> .8997

2.  Or you look up the Z score in the table to find your X and then subtract that number from 1 (one) and that would be the answer.

-1.28 –> .1003       1 – .1003 = .8997

*Moral of the story, whichever way you feel comfortable using, you will get the same answer.

Step 5: Check back to see what the problem was asking and see if we answered it. In this case we did, so the problem is now complete.

Answer: .8997






Dania Elder

Problem 20 :

How many 5-digit ZIP codes numbers are possible if consecutive digits must be different?

  • First by reading the problem you see there are 5 possible spaces to fill                    _ _ _ _ _
  • Next you see that consecutive digits must be different. Therefore you cannot have a zip code such as 11234, because the two 1’s are consecutive with one another.
  • So for the first spot you have 10 possible outcomes (0-9) as a zip code possibility.  10 _ _ _ _
  • The next spot cannot be the same number as the first spot so there are 9 possible out comes

10 9 _ _ _

  • And the same for the next three spaces.

10 9 9 9 9

  • Lastly, you Multiply the possible outcomes.

10*9*9*9*9 = 65,610 

Study Guide for Problem #12

Problem #(12), By Anil K. Dipu

The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of 4.5

a)    Find the probability that in a given year there will be less than 21 earthquakes.

b)    Find the probability that in a given year there will be between 18 and 23 earthquakes.

Step 1. Part (a)  List all the given facts for the first statement.


The statement says that the situation is approximately normally distributed for the number of earthquakes in a year.

This simply means that we can use the “Standard Normal Distribution Curve” for µ=0 and make it equal to the mean that it gives us to use from the statement which is 20.8 as the new µ (“Miu”-mean). Thus the curve is balanced on both sides. (see example below)

The mean is now 20.8, (µ=20.8).

The standard deviation sigma, is equal to 4.5 (σ=4.5) which we will use to find the Probability with the mean (µ=20.8)

Step 2. Analyze problem (a). And draw a standard Normal distribution curve. And label the µ and X.

In both a and b, we do not have an exact sample of population “n”, or success “p” so we do not have to use the subtraction or addition of 0.5.

Therefore we can use the normal method to find the probability with the Z score from the Standard distribution Table which only gives the probability as an area to the left of Z and X.

“Z” represents a result for a part of “1” under the normal standard distribution curve.

The “X” is a piece of data that relates to an area or fraction under the curve. It belongs to entire data series that can be near or away from the mean (µ). Though X is not area, it is a continuous a number.

For problem a, the probability must be less than the X of 2. Therefore X can be understood as less than 21

(X= less than 21).  The probability is not X and it is not 21. It is an area under the curve which is less than 21.

To find the area that of X= less than 21 we draw a standard distribution curve and label the Miu which is 20.8 as the middle or center of the data.  We mark where x is located.

As you will see, 21 is a bit further on the right side of (µ=20.8), but we need an area that is less than 21.

Since we are using a table which shows areas to the left of Z and X we have to shade or mark the area that is less than 21. The area will be on the left.

This also means that we Do Not have to subtract the final value of area from 1 since the area that is already on the left is less than the X of 21.

Step 3. Find Z and Area for probability.

We now use the regular method to find Z in order to find the value of Probability as an area which means a X=probability is less than 21.

Z= (X- µ)/ σ  = (21-20.8)/ 4.5= 0.2/4.5 =  0.0444444444444444 =0.4

(round to the nearest two decimal places after the “point or period” for z since the table uses a few decimals)

(A positive, “+” number will mean that is an area of Z which is greater than the Miu for 20.8).

Step 4. Find value of Probability with the Z score

The table that we are going to use is Table 4. We can see that 0.04 of z is an area of .5160 (=.516) which is understood as a Probability less than 21 which also on the left under the curve for normal and standard distribution.

Step 5. Part (b) List all the given facts for the first statement.


Standard Deviation is σ =4.5 and mean is µ=20.8

X is 18 and 23. The probability is supposed to be between these two data points.

Step 6. Draw the standard distribution table with the Miu and two X’s labeled.

From the standard normal distribution curve we can see that the area is between the two data points for x of 18 and 23. Therefore we need to try find the value of that area.

(Note 1: That part b did not say less than or more than so still do not need to worry about subtracting our final answer from 1, Note 2: Also since the table will always show an area that is to the left for X and Z, we are only looking at areas that concern the middle portion between the two X’s of 18 and 23. With corresponding Z’s)

Step 7. Find the two z scores, with their corresponding areas in order to find the actual difference for area of Probability.

For the first Z we use the first x of 18 and subtract Miu of 20.8, then by standard deviation 4.5.

We will now proceed as follows (18 – 20.8)/4.5 = -.062, with area of 0.2672 for A1

(For a negative value of Z, use Table 1 for area of Z less than Miu)

The second Z is the same but we use 23 for the second X.

The process will be written as (23 – 20.8)/4.5 = .49, with area of 0.6879 for  A2

Step 8. Find the probability between the two X’s of 18 and 23.

The method for probability between two values of X will be stated as follows: A2-A1

(The larger Area subtract the smaller Area)

Then we can use the table which corresponds to the new Z to find the Probability of area under the curve.

The new area is calculated by 0.6879 – 0.2676 = 0.4203

Step 9. Review, Study, & Leave A Comment if preferred.