Professional blog

Sheet1

Year
Make
Model
Price ($)
MPG/C
MPG/H
Midsize
Annual maintenance ($)

2024
Toyota
Camry
25000
28
39
Midsize
388

Honda
Accord
24500
30
38
Midsize
400

Hyaunda
Sonata
24000
28
38
Midsize
458

Nissan
Altima
24300
28
39
Midsize
483

Ford
Fusion
23170
23
34
Midsize
581

Chevolett
malibu
23000
29
36
Midsize
532

Mazda
6
24100
26
35
Midsize
481

Kia
Optima
23390
24
32
Midsize
639

Subaru
Legacy
23475
27
35
Midsize
471

Volkswagon
Passat
22995
24
36
Midsize
563

Bugatti
Chiron
3,900,000

Descriptive Statistics

Mean
23793.00

Standard Error
217.77

Column1

Median
23737.50

Mode
ERROR:#N/A

Mean
376175.454545455

Standard Deviation
688.650854933

Standard Error
352382.509601334

Sample Variance
474240

Median
24000

Kurtosis
-1.0036452782

Mode
ERROR:#N/A

Skewness
0.3970704861

Standard Deviation
1168720.56703143

Range
2005

Sample Variance
1365907763802.27

Minimum
22995

Kurtosis
10.9999908339

Maximum
25000

Skewness
3.3166228904

Sum
237930

Range
3877005

Count
10

Minimum
22995

Maximum
3900000

Sum
4137930

Count
11

Sheet2

Sheet3

1

Descriptive Statistics

Student Name

Institutional Affiliation

Course

Instructor

Date

2

Descriptive Statistics

Descriptive Statistics

Mean 23793.00

Standard Deviation 688.65

Median 23737.50

From the above data, the mean of the data is 23,793.00, standard deviation is 688.65, and

the median of the data is 23,737.50. Therefore, most of the data point is close to the mean as

indicated by the small value of standard deviation (Livingston, 2004). Therefore, the prices of

the cars are not widely varied. The mean is the average of the observations in the dataset, while

the median is the middle most data point in the distribution. Since both the median and the mean

are close to each other the prices of the different cars can be said to be symmetrically distributed.

Column

Mean 376175.45

Median 24000.00

Standard Deviation 1168720.57

The addition of an eleventh car that costs at least one million has the largest effect on the

standard deviation. The variation of the observations from the mean was 1,168,720.57, which

shows a great variability that indicated the data is not normally distributed (Livingston, 2004).

3

Additionally, the mean and the median were not close to each other, which indicated

asymmetrical distribution of the data. The was surprising in that addition of only one outlier

changed the entire outlook of the data.

4

References

Livingston, E. H. (2004). The mean and standard deviation: what does it all mean?. Journal of

Surgical Research, 119(2), 117-123.

Binomial Probabilities

The Probability of an event can be expressed as a binomial probability if its

outcomes can be broken down into two probabilities, p, which is a success and a

q, which is a failure. Where p and q are complementary

p + q = 1, thus q = 1 – p

You need to rewrite the probabilities in the less than or equal to form to use the

function in EXCEL. We will use Excel to find Binomial Probabilities. The

probabilities do need to be in the less than or equal to form to use Excel. This is

very important.

Here are some common Binomial Probabilities and how they would get re-written

to calculate in the less than or equal to form, to use Excel.

• P( x = j) • P( x ≤ j) • P( x < j) = P(x ≤ j – 1) • P(x > j) = 1 – P(x ≤ j) • P( x ≥ j) same as 1 – P(x ≤ j – 1) • Expected Value = n*p

• Standard deviation = √𝑛 ∗ 𝑝 ∗ 𝑞

• Recall, “n” denotes the sample size To find Binomial Probabilities we will use the =BINOM.DIST( ) function.

Let’s use our Car Price Data from Week 2 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 found the average = $25,650.

Looking at our data we see that 7 out of 10 cars fall below the average. We will

call this a success if your price falls below the average. This means p = 7/10 = .70

and q = 1 – .70 = .30.

If you were to find another random sample of 10 cars based on the same data,

what is the probability that exactly 5 of them will fall below the average?

Because of the word “exactly” we want to find this probability P(x = 5). We will

use the BINOM.DIST() function to find this probability.

P(x = 5) = BINOM.DIST(5, 10, .70, FALSE)

In Excel make sure you hit the “=“ sign first then start typing in BINOM.DIST(

From here make sure you include the left parenthesis then type in the x value,

the n value, the p value (the probability), then either TRUE or FALSE. Then close

the parenthesis ) and hit Enter.

Type in a TRUE when you have a less than or equal to probability and type in a

FALSE when you have an equals probability. This example has an “=“ sign so we

will use a FALSE.

There is a 10.29% probability that exactly 5 of the cars will fall below the average.

Note: When you hit “Enter” the answer will return as a decimal, .1029. You will

then need to convert it to a percent.

2) If you were to find another random sample of 10 cars based on the same data,

what is the probability that fewer than 8 of them will fall below the average?

Because of the word “fewer” we will use the less than sign.

This is the probability we want to find, P(x < 8)

This probability is in the less than form NOT the less than or equal to form so we

need to rewrite this in the less than or equal to form.

Remember: P( x < j) = P( x ≤ j – 1)

P( x < 8) = P(x ≤ 8 – 1) = P(x ≤ 7). Now that the probability is in the less than or

equal to form we can use Excel.

P(x ≤ 7) = BINOM.DIST(7,10,.70,TRUE)

In Excel make sure you hit the “=“ sign first then start typing in BINOM.DIST(.

From here make sure you include the left parenthesis then type in the x value,

the n value, the p value (the probability), then either TRUE or FALSE. Then close

the parenthesis ) and hit Enter.

Type in a TRUE when you have a less than or equal to probability and type in a

FALSE when you have an equals probability. This example has an “≤“ sign so we

will use a TRUE.

There is a 61.72% probability that fewer than 8 of the cars will fall below the

average.

Note: When you hit “Enter” the answer will return as a decimal, .6172. You will

then need to convert it to a percent.

3) If you were to find another random sample of 10 cars based on the same data,

what is the probability that at least 3 of them will fall below the average?

Because of the words “at least” we will use the greater than or equal to sign.

This is the probability we want to find, P(x ≥ 3).

This probability is in the greater than or equal to form NOT the less than or

equal to form so we need to rewrite this in the less than or equal to form.

Remember: P( x ≥ j) = 1 – P( x ≤ j – 1)

P( x ≥ 3) = 1 – P(x ≤ 3 – 1) = 1 – P(x ≤ 2). Now that the probability is in the less than

or equal to form we can use Excel.

1 – P(x ≤ 2) = 1- BINOM.DIST(2,10,.70,TRUE)

In Excel make sure you hit the “=“ sign first, then the 1 – and then, BINOM.DIST(.

From here make sure you include the left parenthesis then type in the x value,

the n value, the p value (the probability), then either TRUE or FALSE. Then close

the parenthesis ) and hit Enter.

Type in a TRUE when you have a less than or equal to probability and type in a

FALSE when you have an equals probability. This example has an “≤“ sign so we

will use a TRUE.

There is a 99.84% probability that at least 3 of the cars will fall below the average.

Note: When you hit “Enter” the answer will return as a decimal, .9984. You will

then need to convert it to a percent.

4) If you were to find another random sample of 10 cars based on the same data,

what is the probability that more than 5 of them will fall below the average?

Because of the word “more” we will use the greater than sign.

This is the probability we want to find, P(x > 5).

This probability is in the greater than form NOT the less than or equal to form so

we need to rewrite this in the less than or equal to form.

Remember: P(x > j) = 1 – P(x ≤ j)

P( x > 5) = 1 – P(x ≤ 5)= 1 – P(x ≤ 5). Now that the probability is in the less than or

equal to form we can use Excel.

1 – P(x ≤ 5) = 1- BINOM.DIST(5,10,.70,TRUE)

In Excel make sure you hit the “=“ sign first, then the 1 – and then, BINOM.DIST(.

From here make sure you include the left parenthesis then type in the x value,

the n value, the p value (the probability), then either TRUE or FALSE. Then close

the parenthesis ) and hit Enter.

Type in a TRUE when you have a less than or equal to probability and type in a

FALSE when you have an equals probability. This example has an “≤“ sign so we

will use a TRUE.

There is a 84.97% probability that more than 5 of the cars will fall below the

average.

Note: When you hit “Enter” the answer will return as a decimal, .8497. You will

then need to convert it to a percent.

Poisson Probabilities

The Poisson probability distribution gives the probability of a number of events

occurring in a fixed interval of time or space if these events happen with a known

average rate and independently of the time since the last event.

You will need to make sure you are looking at the correct time or interval when

you calculate Poisson probabilities. For example, if you can type, on average, 50

words in 1 minute, but you want to calculate a probability for a time of 3 minutes,

then the new average will be 50*3 = 150 words. You would use 150 to calculate

the probability NOT 50. We will do more examples below.

You need to rewrite the probabilities in the less than or equal to form to use the

function in EXCEL. We will use Excel to find Poisson Probabilities. The

probabilities do need to be in the less than or equal to form to use Excel. This is

very important.

Here are some common Poisson Probabilities and how they would get re-written

to calculate in the less than or equal to form, to use Excel.

• P( x = r) • P( x ≤ r) • P( x < r) = P(x ≤ r – 1) • P(x > r) = 1 – P(x ≤ r) • P( x ≥ r) same as 1 – P(x ≤ r – 1) • Expected Value is µ or 𝜆 • r is the number of occurrences

To find Poisson Probabilities we will use the =POISSON.DIST( ) function.

Example:

A Cognitive scientific designed an experiment to measure x, the number of a

times a reader’s eye fixated on a single word before moving past that word in 1

minute. They found that a person’s eye fixated on a word 3 times in 1 minute

before moving on to the next word.

1) Find the probability that a person’s eye will fixated on a word exactly 5 times in

1 minute?

Because of the word “exactly” we want to find this probability P(x = 5). We will

use the POISSON.DIST() function to find this probability.

P(x = 5) = POISSON.DIST(5,3, FALSE)

In Excel make sure you hit the “=“ sign first then start typing in POISSON.DIST(

From here make sure you include the left parenthesis then type in the x value,

then the mean, then either TRUE or FALSE. Then close the parenthesis ) and hit

Enter.

Type in a TRUE when you have a less than or equal to probability and type in a

FALSE when you have an equals probability. This example has an “=“ sign so we

will use a FALSE.

There is a 10.08% probability that a person’s eye will fixate on a word exactly 5

times before it moves on to the next word.

Note: When you hit “Enter” the answer will return as a decimal, .1008. You will

then need to convert it to a percent.

2) Find the probability that a person’s eye will fixate on certain words fewer than

100 times in 30 minutes?

The interval we want to look at for this problem is 30 minutes NOT 1 minute.

Because of this, we need to find the new mean. If a person’s eye fixates on a

word 3 times in 1 minute, then 3*30 = 90. Then a person’s eye will fixate on

certain words 90 times in a 30-minute time period. The new average is 90. This is

the value we will use in the Excel function.

Because of the word “fewer” we will use the less than sign.

This is the probability we want to find, P(x < 100)

This probability is in the less than form NOT the less than or equal to form so we

need to rewrite this in the less than or equal to form.

Remember: P( x < r) = P( x ≤ r – 1)

P( x < 100) = P(x ≤ 100 – 1) = P(x ≤ 99). Now that the probability is in the less than

or equal to form we can use Excel.

P(x ≤ 99) = POISSON.DIST(99,90,TRUE)

In Excel make sure you hit the “=“ sign first then start typing in POISSON.DIST(.

From here make sure you include the left parenthesis then type in the x value,

then the mean, then either TRUE or FALSE. Then close the parenthesis ) and hit

Enter.

Type in a TRUE when you have a less than or equal to probability and type in a

FALSE when you have an equals probability. This example has an “≤“ sign so we

will use a TRUE.

There is an 84.18% probability that a person’s eye will fixate on certain words

fewer than 100 times in a 30-minute period before it moves on to the next word.

Note: When you hit “Enter” the answer will return as a decimal, .8418. You will

then need to convert it to a percent.

3) Find the probability that a person’s eye will fixated on certain words at least 95

times in 30 minutes?

The time frame we are looking at is still 30 minutes so the mean will be 90.

Because of the words “at least” we will use the greater than or equal to sign.

This is the probability we want to find, P(x ≥ 95).

This probability is in the greater than or equal to form NOT the less than or

equal to form so we need to rewrite this in the less than or equal to form.

Remember: P( x ≥ r) = 1 – P( x ≤ r – 1)

P( x ≥ 95) = 1 – P(x ≤ 95 – 1) = 1 – P(x ≤ 94). Now that the probability is in the less

than or equal to form we can use Excel.

1 – P(x ≤ 94) = 1- POISSON.DIST(94, 90,TRUE)

In Excel make sure you hit the “=“ sign first, then the 1 – and then,

POISSON.DIST(. From here make sure you include the left parenthesis then type

in the x value, then the mean, then either TRUE or FALS

The post Recall the car data set you identified in Week 2.? You will want to calculate the average for your data set.? (Be sure you use the numbers without the supercar outlier)? Once you have the first appeared on Writeden.