Tag Archives: Z Score

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

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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 withĀ Ā million andĀ million. Ā 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:Ā 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 inĀ Step 1, so now to plug those numbers in for each of our x’s:

1.Ā z = 150-134/20

z= 16/20

z = 0.8

2.Ā z =Ā 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 isĀ .7881 and the area for the z-score -1.7 isĀ .0446.

 

Step 4.

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

.7881–.0446 =Ā .7435

So the answer toĀ our problem is:

TheĀ 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! šŸ™‚

STUDY GUIDE FOR PROBLEM # 15

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STUDY GUIDE FOR PROBLEM # 15

By Anthony Marc

 

15) It is claimed that the average annual per person spending on prescription drugs is $410. If a survey of 65 randomly selected people indicated an average spending of $425 with a standard deviation of $45, do we reject the claim that the average is $410? Use a 5% level of significance.

  • The first thing to do is construct a hypothesis test with a null and alternative hypothesis to test whether the claim will be greater, less than or equal to the average of what is already stated. The tails of the test refer to the area in the normal curve which corresponds to the alternative hypothesis (the region where we you reject the null hypothesis). The claim states that the average per person annually spending on prescription drugs is $410.
  • Ho (The null hypothesis) would be that the average spent would be equal to $410 and Ha (the alternative hypothesis) would also not be equal to $410. Both hypotheses are considered not equal due to the fact that average is NOT more or less than $410 but is exactly that number $410 being spent.
  • Itā€™s a two tailed test
  • The level of significance is the amount on the normal curve whose maximum probability determines whether or not to reject the claim from the mean to the rejection area.
  • The level of significance is 5% which comes out to 5/100= .0500. Since the normal distribution curve is two tailed the level of significance has to be divided by two to determine the rejection area.
  • .0500/2= .0250 for both ends of the curve. Looking on the table the critical value comes out to z = -1.96.

Ā 

Ā  SOLUTION:

  • The problem would be done as: ,Ā  Z= 425-410/45 and the square root of 65
  • What made it easier for me to solve this problem was do it step by step starting off with the square root of n= 65, which comes out to 8.062257748.
  • Then divide that sum by the standard deviation of 45, so you would divide 45/8.062257748 which comes to the answer of 5.581563057.
  • Then subtract the average by the sample mean (425-410) and divide (5.581563057) which come to the answer of Z= 2.687419249.
  • So as a result we REJECT the claim.

The claim is rejected due to the fact that the Z score is higher than the critical value which determines whether or not it falls within the range of values whether the claim is acceptable. So people annually spend more than the averageĀ $410 per year on prescription drugs.

 

 

Study Guide for Problem #12

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

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

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

Given:

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 ā€œStandard Normal Distribution Curveā€ for Āµ=0 and make it equal to the mean that it gives us to use from the statement which is 20.8 as the new Āµ (ā€œMiuā€-mean). Thus the curve is balanced on both sides. (see example below)

The mean is now 20.8, (Āµ=20.8).

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

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

In both a and b, we do not have an exact sample of population ā€œnā€, or success ā€œpā€ 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.

ā€œZā€ represents a result for a part of ā€œ1ā€ under the normal standard distribution curve.

The ā€œXā€ 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 (Āµ). 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).Ā  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.Ā  We mark where x is located.

As you will see, 21 is a bit further on the right side of (Āµ=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- Āµ)/ ĻƒĀ  = (21-20.8)/ 4.5= 0.2/4.5 = Ā 0.0444444444444444 =0.4

(round to the nearest two decimal places after the ā€œpoint or periodā€ for z since the table uses a few decimals)

(A positive, ā€œ+ā€ 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.

Given:

Standard Deviation is Ļƒ =4.5 and mean is Āµ=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 Xā€™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 Xā€™s of 18 and 23. With corresponding Zā€™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 ā€“ 20.8)/4.5 = .49, with area of 0.6879 for Ā A2

Step 8. Find the probability between the two Xā€™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.

Thankyou,

Anil