Question

There are 10 employees in a particular division of a company. Their salaries have a mean of $70,000, a median of $55,000, and a standard deviation

Answer 1

Answer:

A) $bar{x}_{new}=160,000$

B) Median remains the same.

C) $sigma_{new}=300998.34$

Step-by-step explanation:

Consider the complete question attached below.

No. of employees = n = 10

Given mean = $70,000

Median = $55,000

Standard deviation = $60,000

Largest number on the list = $100,000

Accidentally changed to = $1,000,000

Modified mean, media, SD =?

A) Modified Mean:

$\bar{x}=\frac{\sum x}{n} = 70,000\\\\\sum x=(\bar{x})(n) = (70,000)(10)\\\\\sum x=700,000\\\\\sum x_{new} =700,000 -100,000+1,000,000\\\\\sum x_{new}=1,600,000\\\\\bar{x}_{new}=\frac{1,600,000}{10}\\\\\bar{x}_{new}=160,000$

B) Modified Median:

Median remains same and is not affected by changing highest value.

C) Modified SD:

Standard deviation is given by formula:

$\sigma=\sqrt{\frac{\sum x^{2}-n\bar{x}}{N-1}}---(1)\\\\\sigma_{new}=\sqrt{\frac{\sum x_{new}^{2}-n\bar{x}_{new}}{N-1}}---(2)\\\\From\,\, (1)\\\\\sum x^{2}=(N-1)\sigma^{2}+n\bar{x}$

$\sum x^{2}=(10-1)(60,000)^{2}+(10)(70,000)^{2}\\\\\sum x^{2}=8.14\times 10^{10}\\\\\sum x_{new}^{2}= 8.14\times 10^{10}-(10,0000)^{2}+(1,000,000)^{2}\\\\\sum x_{new}^{2}=1.0714\times 10^{12}\\\\Using\,\, (2)\\\sigma_{new}=\sqrt{\frac{1}{9}(1.0714\times 10^{12}-(10)(1.6\times 10^{5})}\\\\\sigma_{new}=\sqrt{9.06\times 10^{11}}\\\\\sigma_{new}=300998.34$