Question

Teacher raises A school system employs teachers at

Answer 1

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) The shape of the data remains the same

(b) The mean and median are increased by $1,000

(c) The standard deviation and interquartile range remain the same

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;

$\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}$

Therefore, following an increase of $1,000, we have;

 $\overline x_{New} = \dfrac{\sum (f_i \cdot (x_i + 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i + f_i \cdot 1000))}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i}$

$\overline x_{New} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + \dfrac{\sum (f_i \cdot 1000)}{\sum f_i} = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i} + 1000 = \overline x + 1000$

• Therefore, the new mean, is equal to the initial mean increased by 1,000

Median;

Given that all salaries, $x_i$, are increased by $1,000, the median salary, $x_{med}$, 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 $\sigma =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}}$;

Where;

n = The number of teaches;

Given that, we have both a salary, $x_i$, and the mean, $\overline x$, increased by $1,000, we can write;

$\sigma_{new} =\sqrt{\dfrac{\sum \left ((x_i + 1000) -(\overline x + 1000)\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 -\overline x - 1000\right )^{2} }{n}} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}}$

$\sigma_{new} = \sqrt{\dfrac{\sum \left (x_i + 1000 - 1000 - \overline x\right )^{2} }{n}} =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}} = \sigma$

Therefore;

$\sigma_{new} = \sigma$; The standard deviation stays the same

Interquartile range;

The interquartile range, IQR = Q₃ - Q₁

New interquartile range, IQR$_{new}$ = (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR

Therefore;

• The interquartile range stays the same

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