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

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

B) Median remains the same.

C)
`sigma_(new)=300998.34`

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}=(\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)=(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* 10^(10)\\\\\sum x_(new)^(2)= 8.14* 10^(10)-(10,0000)^(2)+(1,000,000)^(2)\\\\\sum x_(new)^(2)=1.0714* 10^(12)\\\\Using\,\, (2)\\\sigma_(new)=\sqrt{(1)/(9)(1.0714* 10^(12)-(10)(1.6* 10^(5))}\\\\\sigma_(new)=\sqrt{9.06* 10^(11)}\\\\\sigma_(new)=300998.34`