By adding a constant value to every salary amount, the measures of
central tendency are increased by the amount, while the measures of
dispersion, remains the same
The correct responses are;
(a) <u>The shape of the data remains the same</u>
(b) <u>The mean and median are increased by $1,000</u>
(c) <u>The standard deviation and interquartile range remain the same</u>
Reasons:
The given parameters are;
Present teachers salary = Between $38,000 and $70,000
Amount of raise given to every teacher = $1,000
Required:
Effect of the raise on the following characteristics of the data
(a) Effect on the shape of distribution
The outline shape of the distribution will the same but higher by $1,000
(b) The mean of the data is given as follows;
![\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}](https://tex.z-dn.net/?f=%5Coverline%20x%20%3D%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%20x_i%29%7D%7B%5Csum%20f_i%7D)
Therefore, following an increase of $1,000, we have;
![\overline x_{New} = \dfrac{\sum (f_i \cdot (x_i + 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i + f_i \cdot 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i}](https://tex.z-dn.net/?f=%5Coverline%20x_%7BNew%7D%20%3D%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%20%28x_i%20%2B%201000%29%29%7D%7B%5Csum%20f_i%7D%20%3D%20%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%20x_i%20%2B%20f_i%20%5Ccdot%201000%29%29%7D%7B%5Csum%20f_i%7D%20%3D%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%20x_i%29%7D%7B%5Csum%20f_i%7D%20%2B%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%201000%29%7D%7B%5Csum%20f_i%7D)
![\overline x_{New} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + 1000 = \overline x + 1000](https://tex.z-dn.net/?f=%5Coverline%20x_%7BNew%7D%20%3D%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%20x_i%29%7D%7B%5Csum%20f_i%7D%20%2B%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%201000%29%7D%7B%5Csum%20f_i%7D%20%3D%20%5Cdfrac%7B%5Csum%20%28f_i%20%5Ccdot%20x_i%29%7D%7B%5Csum%20f_i%7D%20%2B%201000%20%3D%20%5Coverline%20x%20%2B%201000)
- Therefore, the new mean, is equal to the initial mean increased by 1,000
Median;
Given that all salaries,
, are increased by $1,000, the median salary,
, is also increased by $1,000
Therefore;
- The correct response is that the median is increased by $1,000
(c) The standard deviation, σ, is given by
;
Where;
n = The number of teaches;
Given that, we have both a salary,
, and the mean,
, increased by $1,000, we can write;
![\sigma_{new} =\sqrt{\dfrac{\sum \left ((x_i + 1000) -(\overline x + 1000)\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}}](https://tex.z-dn.net/?f=%5Csigma_%7Bnew%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Csum%20%5Cleft%20%28%28x_i%20%2B%201000%29%20-%28%5Coverline%20x%20%20%2B%201000%29%5Cright%20%29%5E%7B2%7D%20%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B%5Csum%20%5Cleft%20%28x_i%20%2B%201000%20-%5Coverline%20x%20%20-%201000%5Cright%20%29%5E%7B2%7D%20%7D%7Bn%7D%7D)
![\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}}](https://tex.z-dn.net/?f=%5Csigma_%7Bnew%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B%5Csum%20%5Cleft%20%28x_i%20%2B%201000%20-%5Coverline%20x%20%20-%201000%5Cright%20%29%5E%7B2%7D%20%7D%7Bn%7D%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B%5Csum%20%5Cleft%20%28x_i%20%2B%201000%20-%201000%20-%20%5Coverline%20x%5Cright%20%29%5E%7B2%7D%20%7D%7Bn%7D%7D)
![\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}} =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}} = \sigma](https://tex.z-dn.net/?f=%5Csigma_%7Bnew%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B%5Csum%20%5Cleft%20%28x_i%20%2B%201000%20-%201000%20-%20%5Coverline%20x%5Cright%20%29%5E%7B2%7D%20%7D%7Bn%7D%7D%20%3D%5Csqrt%7B%5Cdfrac%7B%5Csum%20%5Cleft%20%28x_i-%5Coverline%20x%20%20%5Cright%20%29%5E%7B2%7D%20%7D%7Bn%7D%7D%20%3D%20%5Csigma)
Therefore;
; <u>The standard deviation stays the same</u>
Interquartile range;
The interquartile range, IQR = Q₃ - Q₁
New interquartile range, IQR
= (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR
Therefore;
- <u>The interquartile range stays the same</u>
Learn more here:
brainly.com/question/9995782