Question

Prices for used stainless steel side trim for a 1957 chevrolet convertible are $350, $400, $500, $650, $725, $850, and $1700

A.Find the mean to the nearest dollar
B.Find the median
C.Find the four quartiles
D.Find the range(IQR=Q3-Q1)
E. Find the boundary for the lower outliers (Q1-1.5(IQR)).Are there any lower outliers?
F. Find the boundary for the upper outliers(Q3+1.5(IQR)).Are there any upper outliers?

Answer 1

Step-by-step explanation:

here are step-by-step explanations for each part of the problem:

A. To find the mean, you add up all of the prices and divide by the number of prices. So:

Mean = (350 + 400 + 500 + 650 + 725 + 850 + 1700) / 7

Mean = 517.86

Rounded to the nearest dollar, the mean is $518.

B. To find the median, you need to put the prices in order from smallest to largest and then find the middle value. In this case, there are seven prices, so the middle value is between the 3rd and 4th prices:

350, 400, 500, 650, 725, 850, 1700

The median is the average of the 3rd and 4th prices, so:

Median = (500 + 650) / 2

Median = 575

The median price is $575.

C. Quartiles divide the data set into four equal parts, so you need to find the values that separate the lowest 25%, the next 25%, the next 25%, and the highest 25% of prices.

To find the quartiles, you can use the following steps:

Put the prices in order from smallest to largest:

350, 400, 500, 650, 725, 850, 1700

Find the median (Q2) as calculated in part B:

Median = $575

Divide the data set into two halves, below and above the median. If the median is included in both halves, exclude it from both.

Lower half: 350, 400, 500, 575

Upper half: 650, 725, 850, 1700

Find the median (Q1) of the lower half:

Q1 = (400 + 500) / 2

Q1 = 450

Find the median (Q3) of the upper half:

Q3 = (725 + 850) / 2

Q3 = 787.5

The four quartiles are Q1=$450, Q2=$575 (median), and Q3=$788.

D. The range is the difference between the highest and lowest values in the data set. So:

Range = highest value - lowest value

Range = $1700 - $350

Range = $1350

The interquartile range (IQR) is the difference between the third quartile (Q3) and the first quartile (Q1). So:

IQR = Q3 - Q1

IQR = $788 - $450

IQR = $338

F. To find the upper boundary for outliers, you add 1.5 times the IQR to Q3. So:

Upper boundary = Q3 + 1.5(IQR)

Upper boundary = $788 + 1.5($338)

Upper boundary = $1282

There are no values above the upper boundary, so there are no upper outliers.

Answer 2

The mean of the prices is $739, the median is $650, the quartiles are $400 and $850, the range is $1350, there are no lower outliers, and there are no upper outliers.

Step-by-step explanation:

To find the mean of the prices, we add up all the values and divide by the total count. In this case, the total count is 7, so we add up the prices: $350 + $400 + $500 + $650 + $725 + $850 + $1700 = $5175. Then, we divide by 7 to find the mean: $5175 / 7 = $739.29. Rounding to the nearest dollar, the mean is $739.

To find the median, we arrange the prices in ascending order: $350, $400, $500, $650, $725, $850, $1700. Since there are 7 values, the median is the middle value, which is $650.

To find the quartiles, first arrange the prices in ascending order. The first quartile (Q1) is the median of the lower half of the data, which is $400. The third quartile (Q3) is the median of the upper half of the data, which is $850.

The range is the difference between the highest and lowest values. In this case, the range is $1700 - $350 = $1350.

The lower boundary for outliers is given by Q1 - 1.5 * IQR, where IQR is the interquartile range (Q3 - Q1). In this case, IQR = $850 - $400 = $450. So, the lower boundary for outliers is $400 - 1.5 * $450 = $400 - $675 = -$275. Since negative values are not possible in this context, there are no lower outliers.

The upper boundary for outliers is given by Q3 + 1.5 * IQR. In this case, the upper boundary is $850 + 1.5 * $450 = $850 + $675 = $1525. Since there are no values greater than $1525 in the data, there are no upper outliers.