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Question: The output from a statistical software package indicates that the mean and standard deviation of a data set consisting of 200 measurements are \(1,500 and \)300, respectively.

a.What are the units of measurement of the variable of interest? Based on the units, what type of data is this: quantitative or qualitative?

b.What can be said about the number of measurements between \(900 and \)2,100? Between \(600 and \)2,400? Between \(1,200 and \)1,800? Between \(1,500 and \)2,100?

Short Answer

Expert verified
  1. Dollar, Quantitative
  2. 戮, 8/9, 0, 0

Step by step solution

01

Identifying the unit of measurement and the data type 

The units of measurement used here is dollars ($).

Dollars represent money and money is quantifiable. We can count money in numbers. Therefore, the data type is quantitative.

02

Chebyshev’s Rule 

According to the Chebyshev鈥檚 rule, we know that,

  1. Only few measurements fall in the first standard deviation from mean.
  2. 75% measurements fall in the second standard deviation from mean.
  3. 8/9 measurements fall in the third standard deviation from mean.
  4. Any number k > 1, (1-1/k2 ) measurements will fall in k standard deviation.
03

Using the above rules we look at the number of measurements in b 

Mean () = $1500

Standard deviation (s) = $300

Minimum = $900

Maximum = $2100

To use rule 4, we need to find k,

x+ks=Max1500+k(300)=2100k(300)=600k=600300k=2

Now that we have k, we can use rule 4

1-1k21-1221-1434

Therefore, 戮 of the measurements lie in 2 standard deviations.

Minimum = $600

Maximum = $2400

To use rule 4, we need to find k,

x+ks=Max1500+k(300)=2400k(300)=900k=900300k=3

Now that we have k, we can use rule 4

1-1k21-1321-1989

Therefore, 8/9 of the measurements lie in 3 standard deviations.

Minimum = $1200

Maximum = $1800

To use rule 4, we need to find k,

x+ks=Max1500+k(300)=1800k(300)=300k=300300k=1

Therefore, none of the measurements lie in 1 standard deviation.

Minimum = $1500

Maximum = $2100

Because the minimum value here is the mean itself, we cannot say anything about the distribution.

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