Question

27. Find average using the following data.

Data: 456 789 158 394 167 891 264 981 657 912

a. If a distribution was skewed, what would the probability be of finding something 3 standard deviations above the mean? Use one decimal place.

b. Using the data from a, find the IQR. Use one decimal place.

c. Using the data from #27, find the standard deviation. Use two decimal places.

d. Find the 7th observation in the data set and standardize it. Use two decimal places.

Please explain steps :)

Answer 1

27. Find average using the following data.

For the given data, we have

Total sum = 5669

Number of observations = 10

Average = Total sum / number of observations = 5669/10 = 566.9

Average = 566.9

a. If a distribution was skewed, what would the probability be of finding something 3 standard deviations above the mean?

According to Chebyshev’s rule, area between 3 standard deviations from mean is given as below:

k = 3

Area between 3SD = 1 – (1/k^2) = 1 – (1/3^2) = 1 – (1/9) = 1 - 0.111111 = 0.888889

So, area outside 3SD = 1 - 0.888889 = 0.111111

Area above 3SD = 0.111111/2 = 0.055556

Required probability = 0.1

Part b

We have n = 10

First quartile = Q1 = 0.25*n’th obs. = 0.25*10 = 2.5 ≈ 3^rd obs. when data is in an increasing order.

Data in increasing order is given as below:

X
158
167
264
394
456
657
789
891
912
981

So, Q1 = 264

Third quartile = Q3 = 0.75*n = 0.75*10 = 7.5 ≈ 8^th obs. when data is in an increasing order.

So, Q3 = 891

IQR = Q3 – Q1 = 891 – 264 = 627

IQR = 627.0

Part c

Formula for standard deviation is given as below:

SD = Sqrt[∑(X - mean)^2/(n – 1)]

X	(X - mean)^2
456	12298.81
789	49328.41
158	167199.21
394	29894.41
167	159920.01
891	105040.81
264	91748.41
981	171478.81
657	8118.01
912	119094.01
Total	914120.9

SD = Sqrt[∑(X - mean)^2/(n – 1)]

SD = Sqrt[914120.9/(10 – 1)]

SD = Sqrt(101568.9889)

SD = 318.6989

SD = 318.70

Part d

We are given 7^th observation = 264

Z = (X – mean) / SD

Z = (264 - 566.9) / 318.70

Z = -0.950423596

Z = -0.95