I can find out the mean, but not S.D /variance. How can I do this? Please show your steps so that I can understand. Thank you

A class of 20 students takes a test. The score, x, for each student was recorded by the teacher. The results are summarized by

Σ(x-10) = 208 and Σ(x-10)² = 2716

(i) Calculate the standard deviation of the 20 scores.

(ii) Show that Σx² = 8876.

(iii) Two other students took the test later. Their scores were 18 and 16. Find the mean and standard deviation of all 22 scores.

Originally Posted by emily9843
I can find out the mean, but not S.D /variance. How can I do this? Please show your steps so that I can understand. Thank you

A class of 20 students takes a test. The score, x, for each student was recorded by the teacher. The results are summarized by

Σ(x-10) = 208 and Σ(x-10)² = 2716

(i) Calculate the standard deviation of the 20 scores.

(ii) Show that Σx² = 8876.

(iii) Two other students took the test later. Their scores were 18 and 16. Find the mean and standard deviation of all 22 scores.
Hi emily9843!

Can you expand Σ(x-10) = 208 in terms of Σx?
And can you also expand Σ(x-10)² = 2716 in terms of Σx and Σx²?

Which formulas do you have for the standard deviation or variance?

Hi, I found out how to do the SD. I did this by working the mean and sd out for y and then adding 10 for the mean of x. However, I am not sure how to do ii and iii. I'm so confused Thank you

Let's take a look at Σ(x-10) = 208.

If we expand this, we get:

$\displaystyle \sum(x-10) = \sum_{i=1}^{20} (x_i - 10) = (x_1 - 10) + (x_2 - 10) + (x_3 - 10) + ... + (x_{20} - 10)$

Rearranging the terms we get:

$\displaystyle \sum(x-10) = (x_1 + x_2 + x_3 + ... + x_{20}) - (10 + 10 + 10 + ... + 10)$

Now we make it more compact again:

$\displaystyle \sum(x-10) = (\sum_{i=1}^{20} x_i) - 20 \times 10 = \sum x - 200$

So we have:

Σ(x-10) = Σx - 200 = 208

Σx = 208 + 200 = 408

Can you do something similar for Σx²?

Hi Miss EMILY

1. divide 208 by 20 to find 10.4 this is the so called the modified mean of the dataset.
then add 10 which is the number you subtract from every data of the data set and the real mean of the data set is 20.4

2. to find the SD use the following formulae: SD=SQROOT(2716/20 -(10.4)^2) and then SD = sqrroot(135.8-108.16)=sqrroot(27.64)=5.25

unlike the mean which depends on the number that you subtract or you add ( modified mean ) the standard deviation remains the same for the real values of x or (x-k) or (x+k)...therefore
to find the Σχ^2 use the following formula: (SD)^2=(Σχ^2)/20-(20.4)^2. Substitute SD=5.25 and make Σχ^2 the subject of the formula...you must find ...Σχ^2=967.41.

Νοw to find the new mean and SD of the 22 students works as follows.
first use the modified data and subtract from each new data the number 10. therefore (18-10) +(16-10) =14

ADD 14 + Σ(χ-10) to find 222 and divide this number by 22 to find the new modified mean =222/22=10.09
this is the new modified mean for the 22 students and the real mean is 10.09 +10 =20.09

to find the new standard deviation you need to add (18-10)^2 +(16-10)^2 to the old summ 2716 to find 2716+100=2816.

then use the same formula for the standard deviation as above to find SD=SQRROOT(2816/22-(10.09)^2 AND YOU MUST FIND SD=SQRROT(26.1919)=5.117

THIS IS THE NEW STANDARD DEVIATION FOR THE 22 STUDENTS.

GOOD LUCK
MINOAS
Please check and verify all the calculations for any typing mistake.....