Respuesta :
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;
[tex]\overline x = \dfrac{\sum (f_i \cdot x_i)}{\sum f_i}[/tex]
Therefore, following an increase of $1,000, we have;
[tex]\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}[/tex]
[tex]\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[/tex]
- Therefore, the new mean, is equal to the initial mean increased by 1,000
Median;
Given that all salaries, [tex]x_i[/tex], are increased by $1,000, the median salary, [tex]x_{med}[/tex], 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 [tex]\sigma =\sqrt{\dfrac{\sum \left (x_i-\overline x \right )^{2} }{n}}[/tex];
Where;
n = The number of teaches;
Given that, we have both a salary, [tex]x_i[/tex], and the mean, [tex]\overline x[/tex], increased by $1,000, we can write;
[tex]\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}}[/tex]
[tex]\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}}[/tex]
[tex]\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[/tex]
Therefore;
[tex]\sigma_{new} = \sigma[/tex]; The standard deviation stays the same
Interquartile range;
The interquartile range, IQR = Q₃ - Q₁
New interquartile range, IQR[tex]_{new}[/tex] = (Q₃ + 1000) - (Q₁ + 1000) = Q₃ - Q₁ = IQR
Therefore;
- The interquartile range stays the same
Learn more here:
https://brainly.com/question/9995782
