The normal distribution I and II
All posts must 100% original work. no plagiarism. Post results must be provided using Excel. Make sure you interpret your results on a Word Document.
Recall the car data set you identified in Forum 2 excel attached. We know that this data set is normally distributed using the mean and SD you calculated. (Be sure you use the numbers without the supercar outlier)
For the next 4 cars that are sampled, what is the probability that the price will be less than $500 dollars below the mean? Make sure you interpret your results.
Please note: we are given a new sample size, we will need to calculate a new SD. Then, to find the value that is $500 below the mean you will need to take the mean and subtract $500 from it. For example, if the mean is $15,000 then $500 below this would be $14,500. Thus the probability you would want to find is P(x < 14,500).
For the next 4 cars that are sampled, what is the probability that the price will be higher than $1000 dollars above the mean? Make sure you interpret your results. Use the same logic as above. If your mean is $15,000 then $1,000 above is 15,000 + 1,000 = $16,000. Thus the probability you would want to find is P(x > 16,000).
For the next 4 cars that are sampled, what is the probability that the price will be equal to the mean? Make sure you interpret your results. Use the same logic as above.
For the next 4 cars that are sampled, what is the probability that the price will be $1500 within the mean? Make sure you interpret your results. Use the same logic as above.
I encourage you to review the Week 4 normal probabilities PDF at the bottom of the discussion. This will give you a step by step example to follow and show you how to find probabilities using Excel. I also encourage you to review the Week 4 Empirical Rule PDF. This will give you a better understanding on how to utilize the empirical rule.
Sheet1
| Type | Year | Make | Mode | Price | MPG(CITY) | MPG (HIGHWAY) | Weight | ||||||||||||
| variable type: qualitative | variable type: quantitative | variable type: quantitative | |||||||||||||||||
| SUV | 2021 | Mazda | CX-30 | $ | 22 | 9 | 25 | 33 | 3232 | ||||||||||
| compact crossover | Toyota | rav4prime | $ | 29 | 36 | 40 | 5,530 | ||||||||||||
| Chrysler | Voyager | $ | 28 | 19 | 4,330 | ||||||||||||||
| minivan | 2020 | kia | Sedona | $28 | 18 | 24 | 6,085 | ||||||||||||
| Dodge | grand caravan | $29 | 17 | 4,5 | 10 | ||||||||||||||
| passenger wagon | Ford | transit connect | $28,315 | 26 | 3,689 | ||||||||||||||
| Volkswagen | Tigwan | $25,965 | 3,847 | ||||||||||||||||
| 2019 | Sorento | $28, | 11 | 3,810 | |||||||||||||||
| Honda | Odyssey | $32,110 | 4,593 | ||||||||||||||||
| Hyundai | Palisade | $33,665 | 4,284 | ||||||||||||||||
| sports car | Bugatti | La Voiture | $3,250,000 | 14 | 4,400 | ||||||||||||||
| SUMMARY BEFORE ADDING OUTLIER | |||||||||||||||||||
| mean | $28,689 | $22 | $4,391 | ||||||||||||||||
| standard deviation | 2977.7926966873 | 5.5467708324 | 4.8579831206 | 863.2415652643 | |||||||||||||||
| median | $28,725 | $21 | $4,307 | ||||||||||||||||
| SUMMARY AFTER ADDING OUTLIER | |||||||||||||||||||
| mean | $321,536 | $27 | $4,392 | ||||||||||||||||
| $971,266 | $7 | $6 | $819 | ||||||||||||||||
| $28,730 | $19 | $4,330 | |||||||||||||||||
| DISCUSSION | |||||||||||||||||||
| From the initial analysis, the following can be established: The average MPG stands at 22 in the city and at 14 on the highway. This defines the average of all the vehicles selected in the list. Therefore, this figure can be used as a substitute for all the MPG figures respectively. However, the MPG figure for the highway deviates from the central measure of tendency with a certain value. This measure of dispersion is determined by the standard deviation. In the MPG in the city, the standard deviation is given by 5.55. this figure is used in determining the range within which most of the data points are found, defined from the central point of tendency ( | Mean |
Without supercar
| 28689.3 | |||
| Standard Error | 941.6607321347 | ||
| Median | 28725 | ||
| ERROR:#N/A | |||
| Standard Deviation | |||
| Sample Variance | 8867249.34444445 | ||
| Kurtosis | 1.2469446572 | ||
| Skewness | -0.3237289483 | ||
| Range | 10870 | ||
| Minimum | 22795 | ||
| Maximum | 33665 | ||
| Sum | 286893 | ||
| Count | |||
| Confidence Level(95.0%) | 2130.1845701243 |
With Supercar
| 321535.727272727 |
| 292847.665977757 |
| 28730 |
| 971265.828779546 |
| 943357310154.818 |
| 10.9997518649 |
| 3.3165733432 |
| 3227205 |
| 3250000 |
| 3536893 |
| 652505.262278539 |
Thisweek we will discuss the Empirical Rule.
The empirical rule allows you to determine the proximity of the data to the mean.
This only works for bell shape or symmetric distributions.
• The interval that is one standard deviation away contains approximately
68% of the data.
(�̅� ± 1 *SD)
• The interval that is two standard deviations away contains approximately
95% of the data.
(�̅� ± 2 *SD)
• The interval that is three standard deviations away contains approximately
99.7% of the data.
(�̅� ± 3 *SD)
Let’s continue to look at the Data from Week 2.
Car Price:
Observation 1 $ 20,000
Observation 2 $ 25,000
Observation 3 $ 30,000
Observation 4 $ 31,000
Observation 5 $ 22,500
Observation 6 $ 25,000
Observation 7 $ 29,500
Observation 8 $ 24,000
Observation 9 $ 24,500
Observation 10 $ 25,000
Mean: $ 25,650
Median: $ 25,000
SD: $ 3,488.47
Sample Size: 10
Using the Empirical Rule calculate how many data points fall within the 1, 2 and 3
SD’s?
1) First, we will need to calculate each interval.
25, 650 – 3,488.47 = $22,162 -> round to the nearest dollar
25, 650 + 3,488.47 = $29,138 -> round to the nearest dollar
The interval for approximal 68% of the data is ($22,162, $29,138).
But how many data points fall within this interval?
We see that observations, 1, 2, 5, 6, 8 ,9, and 10. 7 of the 10 observations fall
within this interval. That is
7
10
= 70% of the data falls within 1 SD. This is very
close to 68%.
2) We will calculate the next interval.
25, 650 – (2) 3,488.47 = $18,673 -> round to the nearest dollar
25, 650 + (2) 3,488.47 = $32,627 -> round to the nearest dollar
The interval for approximal 95% of the data is ($18,673, $32,627).
But how many data points fall within this interval?
We see that observations, 1, 2, 3, 4 5, 6, 7 8 ,9, and 10. All 10 of the 10
observations fall within this interval. That is
10
10
= 100% of the data falls within 2
SD’s. This is very close to 95%. Since this is a smaller data set see that all the data
points fall within the first 2 observations is not uncommon. We would expect
results like this.
3) But it is still a good idea to calculate the last interval.
25, 650 – (3) 3,488.47 = $15,185 -> round to the nearest dollar
25, 650 + (3) 3,488.47 = $36,115 -> round to the nearest dollar
The interval for approximal 99.7% of the data is ($15,185, $36,115).
But how many data points fall within this interval?
Just like with the last interval all the data points fall within this interval and
because this is a small data set the results are as expected.
There are no data points that fall outside this range. There doesn’t appear to be
any outliers in this data set. We also see that the mean and median are close
together. There isn’t a big difference between the two values. Because of these
explanations, this data appears to be normal and have a normal distribution. The
data set does not seem to be skewed, in either direction.
We can see how the SD’s line up along the x-axis and it creates the bell-shaped
curve.
NormalDistributions are the most common distributions in statistics. If a random
variable X is normally distributed with a mean μ and a standard deviation σ.
X ~ N(μ, σ) ; Z ~ N(0, 1)
Normal distributions are known as “bell-shaped curve”
To find the probabilities of normal distributions using a Normal Distribution Table,
we would start by converting the x values to a standard normal z-curve. The
equation of the z – score;
𝑧 =
𝑥 − 𝜇
𝜎
Nowadays, we do not need to do this conversion to the standard normal
distribution, since Excel does it automatically for us.
Excel can only find Less Than probabilities, therefore it is important to make sure
that your problem is only including the less than inequality (<).
Less Than OR Less Than and Equal To is not as important because normal
probabilities are continuous not discrete. Here are some common Normal
Probabilities and how they would get re-written to calculate in the less than form,
to use Excel.
• P(X ≤ j) same as P( X < j) • P( X ≥ j) same as 1 – P(X < j ) • P(j < X < k) = P(X < k) – P(X < j) • Expected Value = µ (Mean) • Standard deviation = σ (SD)
To find Normal Probabilities we will use the =NORM.DIST( ) function.
The Central Limit Theorem states that given any distribution with a mean μ and a
standard deviation of σ, the sample mean will approach a normal distribution as the
sample size, n, increases. The new mean of the sample mean will equal the old
mean; new μ = old μ and the new standard deviation of the sample mean (this is
also called standard error) will be written as;
𝒏𝒆𝒘 𝒔𝒅 (or standard error) =
𝜎
√𝑛
Let’s use our Car Price Data from Week 1 and calculate 4 different probabilities.
Car Price:
Observation 1 $ 20,000
Observation 2 $ 25,000
Observation 3 $ 30,000
Observation 4 $ 31,000
Observation 5 $ 22,500
Observation 6 $ 25,000
Observation 7 $ 29,500
Observation 8 $ 24,000
Observation 9 $ 24,500
Observation 10 $ 25,000
1. Using our data, we believe that the cost of the type of car we calculated is
normally distributed with a mean of $25,650 and a SD of $3,488.47. Assume that 5
additional cars are randomly sampled, and their prices are recorded. What is the
probability that the sample mean price of the 5 new cars will be less than $24,000?
The probability is already in the less than form, P(�̅� < 24,000), so we do not need to do additional work in Excel to find the probability. We also notice that the new sample size is n = 5. The mean will stay the same, but we will need to calculate a new SD. We will apply the Central Limit Theorem to do this. Remember you need to put in the “=” sign and then we will click on the cell that contains the old SD, and will hit the “ / “ sign and then use the SQRT( ) function and put 5 within the parentheses because the new sample size is 5.
𝒏𝒆𝒘 𝒔𝒅 =
𝜎
√𝑛
=
3488.47
√5
= 1560.09
Next, we want to find this probability P(�̅� < 24,000) and we will use the NORM.DIST() function in Excel to do this. P(�̅� < 24000) = NORM.DIST(24000, 25650, 1560.09, true)
In Excel make sure you hit the “=“ sign first, then start typing in NORM.DIST(. From
here make sure you include the left parenthesis then type in the x value, the mean,
the standard deviation, then either True. Then close the parenthesis ) and hit
Enter. ALWAYS type in a True for continuous probability functions (the normal
distribution is continuous). This example has an “<“ sign so we will use a True.
The probability that the sample mean for the new sample of 5 cars is below
$24,000 is 14.51%.
Remember: Once you hit “Enter” the answer returns a decimal. You need to
convert it to a percentage if you want to read a percentage.
2. Assume that 5 additional cars are randomly sampled, and their prices are
recorded. What is the probability that the sample mean price of the 5 new cars will
be higher than $25,000?
Because of the words “higher than”, we want to find this probability P(�̅� > 25,000).
Since we are using the same data the mean and the new SD will be the same.
Remember the function in Excel are in the less than form. This means we will need
to do an extra step in Excel to get the probability we want.
P(�̅� > 25,000) = 1 – NORM.DIST(25000,25650,1560.09,TRUE)
In Excel make sure you hit the “=“ sign first, then the 1 – and then start typing in
NORM.DIST(. From here make sure you include the left parenthesis then type in the
x value, the mean, the standard deviation, then either True. Then close the
parenthesis ) and hit Enter.
The probability that the sample mean for the new sample of 5 cars is below
$25,000 is 66.15%.
Remember: Once you hit “Enter” the answer returns a decimal. You need to
convert it to a percentage if you want to read a percentage.
3. Assume that 5 additional cars are randomly sampled, and their prices are
recorded. What is the probability that the sample mean price of the 5 new cars will
be between $24,000 and $25,000?
Because of the word “between”, we want to find this probability
P(24000 < �̅�< 25000). Since we are using the same data the mean and the new SD
will be the same. Remember the function in Excel are in the less than form. This
means we will need to do an extra step in Excel to get the probability we want.
P(24000 < �̅� < 25000) = P(�̅� < 25000) – P(�̅� < 24000)
= NORM.DIST(25000, 25650, 1560.09,TRUE) - NORM.DIST(24000,25650,1560.09,TRUE)
In Excel make sure you hit the “=“ sign first, then start typing in NORM.DIST(. From
here make sure you include the left parenthesis then type in the x value, the mean,
the standard deviation, then either True. Then close the parenthesis ), hit the minus
– sign then Repeat and then hit Enter.
The probability that the sample mean for the new sample of 5 cars is between
$24,000 and $25,000 is 19.34%.
Remember: Once you hit “Enter” the answer returns a decimal. You need to
convert it to a percentage if you want to read a percentage.