- This topic has 16 replies, 10 voices, and was last updated 9 years, 9 months ago by .
You must be logged in to reply to this topic.
You must be logged in to reply to this topic.
You must be logged in to reply to this topic.
3. The distribution of grades on a quiz taken by 100 students is given in the table
(a) Compute the mean and standard deviation of this distribution. [Use definition of standard deviation (weighted sum of deviations from mean square).]
(b) Find mean & standard deviation of sample mean for random sample of size 16.
Where is thy table? WHERE ART THOU!
I was wondering the same about the table
I was wondering the same about the table
should a table be created??
grade number
1
…………………………………. this is the table/:
i think we can get the table on excel file, isn’t it?
Afzal is right. 10 students got a grade of 1, 40 a grade of 2, 30 a grade of 3 and 20 a grade of 4. The average grade will be between 2 and 3, but not necessarily 2.5. In fact 260 divided by 100 is 2.6.
E(x)=1(.1)+2(.4)+3(.3)+4(.2)=2.6
(x-u)^2 1 2 3 4
x
P(x) 1/10 4/10 3/10 2/10
a) mean = 1+2+6+8/10 =2.6
b) v[x]= e[x-u)^2 = 3.30
c) sd[x]=.825
hmm for A i got
mean = 2.6
SD= 0.916
x P(x) x*P(x) (x-mu)^2 (x-mu)^2*P(x)
1 0.1 0.1 2.56 0.256
2 0.4 0.8 0.36 0.144
3 0.3 0.9 0.16 0.048
4 0.2 0.8 1.96 0.392
2.6 0.84 var
0.916515139 sd
Almost, SD is 0.917.
What is the variance 0.84?
did anyone do part b? does the mean remain the same ?
You must be logged in to reply to this topic.
Ursula C. Schwerin Library
New York City College of Technology, C.U.N.Y
300 Jay Street, Library Building - 4th Floor
Our goal is to make the OpenLab accessible for all users.