The mean and standard deviation of 7 numbers are 5 and 3 respectively.

 

Calculate

 

a) solved

 

b) the new value of the variance if every number is multiplied by 2 and then 5 is added to it.

 

my attempt :

 

 

[math] \sum x = sum-of-numbers [/math]

 

[math] 5 = \frac{\sum x}{7} [/math]

 

[math] {\sum x} = 35 [/math]

 

so, multiply 2 and add 5

 

[math] 35(2) + 5 = 75 [/math]

 

[math] {\sum x} = 75 [/math]

 

find mean,

 

[math] \frac{75}{8} = 9.375 [/math]

 

[math] \sigma^2 = \frac{\sum x^2}{frequency} - mean^2 [/math]

 

[math] \sigma^2 = \frac{75^2}{8} - 9.375^2 [/math]

 

[math] \sigma^2 \approx 615.2344[/math]

 

where the answer is

 

[math] \sigma^2 = 36 [/math]

 

seems to be i'm not good to use variance yet :/

Hi vastor! Intuitively, we know that the variance describes the spread of the numbers. So, adding 5 to each of your numbers wouldn't affect the variance, while multiplying each of the numbers by 2 would. The rule to use in this case is:

 

[latex]Var(ax)=a^2Var(x)[/latex]

The mean and standard deviation of 7 numbers are 5 and 3 respectively.

 

Calculate

 

a) solved

 

b) the new value of the variance if every number is multiplied by 2 and then 5 is added to it.

 

my attempt :

 

 

[math] \sum x = sum-of-numbers [/math]

 

[math] 5 = \frac{\sum x}{7} [/math]

 

[math] {\sum x} = 35 [/math]

 

so, multiply 2 and add 5

 

[math] 35(2) + 5 = 75 [/math]

 

[math] {\sum x} = 75 [/math]

 

find mean,

 

[math] \frac{75}{8} = 9.375 [/math]

 

[math] \sigma^2 = \frac{\sum x^2}{frequency} - mean^2 [/math]

 

[math] \sigma^2 = \frac{75^2}{8} - 9.375^2 [/math]

 

[math] \sigma^2 \approx 615.2344[/math]

 

where the answer is

 

[math] \sigma^2 = 36 [/math]

 

seems to be i'm not good to use variance yet :/

 

When in doubt it usually helps to fall back to the basic definitions, which in the case of a discrete probability space are:

 

 

[math] \overline X = \displaystyle \dfrac {1}{N} \sum_{n=1}^N x_n[/math]

 

[math] \sigma_x^2 = \displaystyle \dfrac {1}{N} \sum_{n=1}^N (\overline X - x_n)^2 [/math]

 

From this you should be able to easily show that [math] \overline {aX}= a \overline X[/math], [math]\overline {X + b} = {\overline X} + b[/math], [math] \sigma_{ax}^2 = a^2 \sigma_x^2 [/math] and [math]\sigma_{x+b}^2 = \sigma_x^2[/math]

When in doubt it usually helps to fall back to the basic definitions, which in the case of a discrete probability space are:

 

 

[math] \overline X = \displaystyle \dfrac {1}{N} \sum_{n=1}^N x_n[/math]

 

[math] \sigma_x^2 = \displaystyle \dfrac {1}{N} \sum_{n=1}^N (\overline X - x_n)^2 [/math]

 

From this you should be able to easily show that [math] \overline {aX}= a \overline X[/math], [math]\overline {X + b} = {\overline X} + b[/math], [math] \sigma_{ax}^2 = a^2 \sigma_x^2 [/math] and [math]\sigma_{x+b}^2 = \sigma_x^2[/math]

 

lmao, talking about basic, this is hardly possible in a formal high-school education which intuitive and imagination is forgotten, thanks to the ministry of education, they just give a bunch of formula and told us plugin the numbers (i don't know what is in their brain, khan academy teach much better)

 

long story short, they just give me the variance formula as follow

 

[math] \sigma^2 = \frac{\sum x^2}{frequency} - mean^2 [/math]

 

and i don't have a clue of what u write... can u elaborate more, or give link so that i can get this thing into my brain?

 

P.S. talking about gaining intuitive, i think i'm the only student in my classroom that can find median of a grouped number without using the given formula, and yes others just memorize the formula and plug-in the number (there nothing to be proud about, such simple intuitive can be gain easily, i'm in a top class anyway, shame to the text book )

Edited October 2, 201114 yr by Vastor