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Standard Deviation, SM-Bank 017

Dataset 1 has mean \(\bar x_1\) and standard deviation \(\sigma_1\).

Dataset 2 has mean \(\bar x_2\) and standard deviation \(\sigma_2\).

Consider the following statement:  If  \(\bar x_1 < \bar x_2\), then  \(\sigma_1 < \sigma_2\).

Is this statement correct? Explain your answer.   (2 marks)

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\(\text{Standard deviation is a measure of how much members}\)

\(\text{of a data group differ from the mean value of the group.}\)

\(\text{It follows that the relative standard deviation between two}\)

\(\text{datasets is not affected if}\ \ \bar x_1 < \bar x_2\).

\(\text{Therefore, the statement is incorrect.}\)

Show Worked Solution

\(\text{Standard deviation is a measure of how much members}\)

\(\text{of a data group differ from the mean value of the group.}\)

\(\text{It follows that the relative standard deviation between two}\)

\(\text{datasets is not affected if}\ \ \bar x_1 < \bar x_2\).

\(\text{Therefore, the statement is incorrect.}\)

Standard Deviation, SM-Bank 003

In a small business, the seven employees earn the following wages per week:

\(\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970\)

  1.  Calculate the standard deviation for this set of data, giving your answer to one decimal place.  (1 mark)

  2.  Each employee receives a $20 pay increase.

     

     Explain the effect that this increase will this have on the standard deviation?  (2 marks)

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i.    \(191.0\)

ii.    \(\text{All values increase by \$20, but so too does the mean.} \)

\(\text{Therefore the spread about the new mean will not change} \)

\(\text{and therefore the standard deviation will remain the same.} \)

Show Worked Solution

i.    \(\text{By calculator (in Statistics mode:}\)

\(\text{Std Dev}\ = 191.044...  =191.0\ \text{(1 d.p.)} \)


ii.    \(\text{All values increase by \$20, but so too does the mean.} \)

\(\text{Therefore the spread about the new mean will not change} \)

\(\text{and therefore the standard deviation will remain the same.} \)

Standard Deviation, SM-Bank 001

The ages of nine students were recorded in the table below.
 

 

  1. What is the standard deviation, correct to two decimal places?   (2 marks)

  2. Briefly explain what is meant by the term standard deviation.   (1 mark)

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i.    \(\text{By calculator (in Statistics mode):} \)

\(\text{Std Dev}\ = 1.5947… = 1.59\ \ \text{(to 2 d.p.)} \)

 

ii.   \(\text{Standard deviation is a measure of how much members}\)

\(\text{of a data group differ from the mean value of the group.}\)

Show Worked Solution

i.    \(\text{By calculator (in Statistics mode):} \)

\(\text{Std Dev}\ = 1.5947… = 1.59\ \ \text{(to 2 d.p.)} \)

 

ii.   \(\text{Standard deviation is a measure of how much members}\)

\(\text{of a data group differ from the mean value of the group.}\)

Statistics, STD1 S1 2020 HSC 24

  1. The ages in years, of ten people at the local cinema last Saturday afternoon are shown.

\(38 \ \ 25 \ \ 38 \ \ 46 \ \ 55 \ \ 68 \ \ 72 \ \ 55 \ \ 36 \ \ 38\)

  1. The mean of this dataset is 47.1 years.
  2. How many of the ten people were aged between the mean age and the median age?  (2 marks)

  3. On Wednesday, ten people all aged 70 went to this same cinema.
  4. Would the standard deviation of the age dataset from Wednesday be larger than, smaller than or equal to the standard deviation of the age dataset given in part (a)? Briefly explain your answer without performing any calculations.  (2 marks)

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a.     \(1\)

b.    \(\text{Standard deviation is a measure of how much the}\)

\(\text{ages of individuals differ from the mean age of the group.}\)
 

\(\Rightarrow\ \text{Standard deviation of Wednesday’s group would be}\)

\(\text{less as the mean is 70 and everyone’s age is 70.}\)

Show Worked Solution

a.     \(\text{Reorder ages in ascending order:}\)

    \(25, 36, 38, 38, 38, 46, 55 , 55, 68, 72\)

\(\text{Median} = \dfrac{\text{5th + 6th}}{2} = \dfrac{38 + 46}{2} = 42\)

\(\therefore\ \text{People with age between 42 − 47.1 = 1}\)

♦ Mean mark (a) 39%.

 
b.    \(\text{Standard deviation is a measure of how much the}\)

\(\text{ages of individuals differ from the mean age of the group.}\)
 

\(\Rightarrow\ \text{Standard deviation of Wednesday’s group would be}\)

\(\text{less as the mean is 70 and everyone’s age is 70.}\)

♦♦♦ Mean mark (b) 20%.

Statistics, STD2 S3 2017 HSC 29d*

All the students in a class of 30 did a test.

The marks, out of 10, are shown in the dot plot.
 

  1. Find the median test mark.  (1 mark)

  2. The mean test mark is 5.4. The standard deviation of the test marks is 4.22.
  3. Using the dot plot, calculate the percentage of the marks which lie within one standard deviation of the mean.  (2 marks)
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  1. \(6\)
  2. \(\text{43%}\)
Show Worked Solution

i.    \(\text{30 data points}\)

\(\text{Median}\ = \dfrac{\text{15th + 16th}}{2} = \dfrac{4+8}{2} = 6\)
 

♦ Mean mark 50%.

ii.    \(\text{Lower limit}\ = 5.4-4.22 = 1.18\)

\(\text{Upper limit} = 5.4 + 4.22 = 9.62\)

\(\text{% between}\) \(= \dfrac{13}{30} \times 100\)  
  \(= 43.33… \%\)  
  \(=43\%\ \ \text{(nearest %)}\)  
♦♦ Mean mark 34%.