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Statistics, STD2 S2 2020 HSC 28

Consider the following dataset.

`1        5        9         10        15`

Suppose a new value, `x`, is added to this dataset, giving the following.

      `1        5        9         10        15        x`

It is known that  `x`  is greater than 15. It is also known that the difference between the means of the two datasets is equal to ten times the difference between the medians of the two datasets.

Calculate the value of `x`.    (4 marks)

--- 8 WORK AREA LINES (style=lined) ---

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`38`

Show Worked Solution

`text{Dataset 1 : Median} = 9`

Mean mark 53%.

`text{Dataset 2 : Median} = frac{9 + 10}{2} = 9.5`

`:.\ text{Difference between means}`

`= 10 xx (9.5 – 9)`

`= 5`
 

`overset_ x_1 = frac{1 + 5 + 9 + 10 + 15}{5} = 8`
  
`therefore overset_ x_2 = 8 + 5 = 13`
 

`text{Sum of all data points in Dataset 2}`

`= 6 xx 13`

`= 78`
 

`78` `= 1 + 5 + 9 + 10 + 15 + x`
`therefore \ x` `= 78 – 40`
  `= 38`

Filed Under: Measures of Centre and Spread, Summary Statistics - No Graph (Std 2) Tagged With: Band 5, common-content, mc-824-10-Mean, smc-6312-10-Mean, smc-6312-20-Median and Mode, smc-824-20-Median and Mode

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