Answer:
A) [tex]bar{x}_{new}=160,000[/tex]
B) Median remains the same.
C) [tex]sigma_{new}=300998.34[/tex]
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:
[tex]\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[/tex]
B) Modified Median:
Median remains same and is not affected by changing highest value.
C) Modified SD:
Standard deviation is given by formula:
[tex]\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}[/tex]
[tex]\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[/tex]
