Vehicle Type/Class (Qualitative)

Year (Quantitative)

Make (Qualitative)

Model (Qualitative)

Price (Quantitative)

MPG (city) (Quantitative)

MPG (highway) (Quantitative)

Sports Car

2021

Aston Martin

V8 VantageRoadster

$214,320

18 Miles

24 Miles

Sports Car

2020

Lamborghini

Huracan EVO

$294,840.00

13 Miles

18 Miles

Sports Car

2020

Ferrari

488 Pista Spider

$699,950.00

15 Miles

19 Miles

SVU

2021

GMC

Yukon Denali

$81,825.00

14 Miles

19 Miles

SVU

2021

Cadillac

Escalade Sport

$99,015.00

14 Miles

19 Miles

SVU

2018

Mercedes-Benz

GLE 43 AMG 4MATIC Coupe

$67,777.00

17 Miles

22 Miles

Sedan

2021

Lexus

ES 250

$45,430.00

25 Miles

34 Miles

Sedan

2018

Honda

Accord

$21,987.00

30 Miles

38 Miles

Sedan

2020

Audi

A8 L 3.0T

$99,245.00

17 Miles

26 Miles

2021

Bentley

Bentayga

$205,999.00

15 Miles

24 Miles

Week 2

The summary statistics are essential for the price data for the previous ten vehicles. Mean 183038.8 Median 99130 Standard Deviation 201021.6022. Based on my statistics summary, the typical vehicle in the price data has a mean of $183,038.8. The median value or the median of the price data is $99,130. Since the mean is greater than the median, we can say that the price data distribution is positively skewed, meaning that most of the price is on the lower side. In addition, there is a standard deviation of $201,021.60. This is a relatively large standard deviation, so we can conclude that the data is spread out, and most of the price data is far from the mean when adding the 11th vehicle and say it a McLaren Speedtail, which costs $2,250,000. Mean 370944.3636 Median 99245 and Standard Deviation 651737.8502. The summary statistics that were most affected by the outlier are the mean and the standard deviation. Adding an outlier that is way larger than the rest has changed the mean and the standard deviation to increase drastically compared to their values without the outlier. This is not surprising since the standard deviation measures how to spread out the values are from the mean, and the compromise of the data will also increase since it involves summing all the prices and dividing it by the sample size. This week's discussion and assignment were very challenging to me, and I know it is just going to get hard as the week goes on. I hope I explained it as best as I could.

Week3

The average price of vehicles is$183,039

Now let us count how many of these vehicles have less price than the average price.

There are6vehicles having a price less than the average.

p=nx=106=0.6

Therefore, the probability of success (p) is0.6

The probability of failure (q) is given by

q=1p=10.6=0.4

Therefore, the probability of failure (q) is0.4

Recall that thebinomial probability distributionis given by

P(x)=nCxpxqnx

Wherenis the number of trials,xis the favorable outcome andpis the probability of success.

nCxis the number of possible combinations (choosexobjects from a total ofnobjects)

For the given case, we have

p = 0.6

q = 0.4

n = 10

a)If you were to find another random sample of 10 cars based on the same data, what is the probability that exactly 4of them will fall

below the average?

Exactly 4 meansx = 4

P(x=4)=10C40.640.4104P(x=4)=2100.640.46P(x=4)=0.1115

Therefore, the probability thatexactly 4of them will fall below the average is0.1115

This means there is an11.15%chance thatexactly 4of the vehicles will have a price below the average price.

b)If you were to find another random sample of 10 cars based on the same data, what is the probability that fewer than 5of them will fall below the average?

Fewer than 5 meansx = 0, 1, 2, 3, 4

P(x<5)=P(x=0)+P(x=1)+P(x=2)+P(x=3+P(x=4)

P(x=0)=10C00.600.4100P(x=0)=0.00011

P(x=1)=10C10.610.4101P(x=1)=0.00157

P(x=2)=10C20.620.4102P(x=2)=0.01062

P(x=3)=10C30.630.4103P(x=3)=0.04247

We have already calculated the probability for x = 4 in the previous part

P(x<5)=0.00011+0.00157+0.01062+0.04247+0.1115P(x<5)=0.1663

Therefore, the probability thatfewer than 5of them will fall below the average is0.1663

This means there is a16.63%chance thatfewer than 5of the vehicles will have a price below the average price.

c)If you were to find another random sample of 10 cars based on the same data, what is the probability that more than 6of them will fall below the average?

More than 6 meansx = 7, 8, 9, 10

P(x>6)=P(x=7)+P(x=8)+P(x=9)+P(x=10

P(x=7)=10C70.670.4107P(x=7)=0.21499

P(x=8)=10C80.680.4108P(x=8)=0.12093

P(x=9)=10C90.690.4109P(x=9)=0.04031

P(x=10)=10C100.6100.41010P(x=10)=0.00605

P(x>6)=0.21499+0.12093+0.04031+0.00605P(x>6)=0.3823

Therefore, the probability thatmore than 6of them will fall below the average is0.3823

This means there is a38.23%chance thatmore than 6of the vehicles will have a price below the average price.

d)If you were to find another random sample of 10 cars based on the same data, what is the probability that at least 4of them will fall

below the average?

At least 4 means equal to or greater than 4

x = 4, 5, 6, 7, 8, 9, 10

Alternatively, we can find the probabilities forx = 0, 1, 2, 3and subtract the result from 1 since this approach will take fewer calculations.

P(x4)=1P(x<4)

P(x<4)=P(x=0)+P(x=1)+P(x=2)+P(x=3)

We have alreadycalculated the probabilities for x = 0, 1, 2, 3 in the previous parts.

P(x<4)=0.00011+0.00157+0.01062+0.04247P(x<4)=0.05477

Finally, the probability is

P(x4)=1P(x<4)P(x4)=10.05477P(x4)=0.9452

Therefore, the probability thatat least 4of them will fall below the average is0.9452

This means there is a94.52%chance thatat least 4of the vehicles will have a price below the average price.

Image source: geeksforgeeks

Using the data set you collected in Week 1, excluding the supercar outlier, you should have calculated the mean and standard deviation during Week 2 forpricedata.

Along with finding apandqfrom Week 3.

Using this information, calculate two 95% confidence intervals.

For the first interval, you need to calculate a T-confidence interval for the sample population.

You have the mean, standard deviation, and sample size, all you have left to find is the T-critical value and you can calculate the interval.

For the second interval calculate a proportion confidence interval using the proportion of the number of cars that fall below the average.

You have thep,q, andn, all that is left is calculating a Z-critical value,

Make sure you include these values in your post, so your fellow classmates can use them to calculate their own confidence intervals.

Once you calculate the confidence intervals you will need to interpret your interval and explain what this means in words.

Do the confidence intervals surprise you, knowing what you have learned about confidence intervals, proportions, and normal distribution?

Please theWeek 5 Confidence T-Interval Mean and Unknown SD PDFand theWeek 5 Confidence Interval Proportions PDFat the bottom of the discussion.

This will give you a step-by-step example of how to help you calculate this using Excel.

These PDFs will also help you in Quizzes section.

Once you have posted your initial discussion, you must reply to at least two other learner's post.

Each post must be a different topic.

So, you will have your initial post from one topic, your first follow-up post from a different topic, and your second follow-up post from one of the other topics.

Of course, you are more than welcome to respond to more than two learners."

Instructions:Youmust respond to at least 2 other students. Responses may include direct questions.

In your first peer posts, pick another confidence level, i.e. 90%, 99%, 97%, any other confidence level is fine.

Have fun and be creative with it and calculate another T-confidence interval and interpret your results.

Compare your results to that of the initial 95%, how much do they differ?

How useful can this type of information be when you go to buy a new car or even a house?

In your second peer post, pick another confidence level, i.e. 90%, 99%, 97%, any other confidence level is fine.

Have fun and be creative with it and calculate another proportion interval and interpret your results.

Compare your results to that of the initial 95%, how much do they differ?

How useful can this type of information be when you go to buy a new car or even a house?

Make sure you include your data set in your initial post as well.

In this discussion, you should be calculating 4 total confidence intervals.

Two T-confidence intervals and Two proportion confidence intervals.

Two in your initial post and then one in each of your response posts.

Week 5 Confidence T-Intervals Mean and Unknown SD.pdf

Week 5 Confidence Intervals Proportions.pdf