mdeclipse03
New Member
I'm not sure if this is posted in the right place but I could sure use some help with a statistic problem:
Problem:
The random variable x has a normal distribution with u 1000 and o 10.
u = 1,000
o = 10
A. Find the probability that x assumes a value more than 2 standard deviations from its mean. More than 3 standard deviations from u
B. Find the probability that x assumes a value within 1 standard deviation of its mean. Within 2 standard deviation of u
C. Find the value of x that represents the 80th percentile of this distribution. The 10th percentile.
At first I used the formula:
z= x-u/o, plugged in the numbers but it was wrong.
The teacher responded with this:
P(z>2) = .5-.4772 = .0228 & P(Z<-2) =P(Z>2) by symmetry. This is the area of the upper tail plus that of the lower tail.
So, P(Z<-2)+P(Z>2)= 2P(Z>2) = 2(.0228)
So after obtaining that information, I resubmitted using the above formula:
4.94 A) P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (from the back of my book)= 2(.0228) = .0456
P(Z<-3)+P(Z>3)= 2P(Z>3)
.5 - .4987 (number from the back of my book) = 2(.0013) = .0026
B) P(Z<-1)+P(Z>1)= 2P(Z>1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204
P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (number from the back of my book) = 2(.0228) = .0456
C) P(Z<- .8 )+P(Z> .8 )= 2P(Z>.8 )
.5 - .2281 (number from the back of my book) = .2719 = 2(.2719) = .5438
P(Z<-.1)+P(Z>.1)= 2P(Z>.1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204
I'm not sure if these even makes sense to anybody but I've been working on this problem for a week now and I have one more week to get it right. Teacher gives us two weeks to make corrections and I am just about lost. I dislike statistics.
Problem:
The random variable x has a normal distribution with u 1000 and o 10.
u = 1,000
o = 10
A. Find the probability that x assumes a value more than 2 standard deviations from its mean. More than 3 standard deviations from u
B. Find the probability that x assumes a value within 1 standard deviation of its mean. Within 2 standard deviation of u
C. Find the value of x that represents the 80th percentile of this distribution. The 10th percentile.
At first I used the formula:
z= x-u/o, plugged in the numbers but it was wrong.
The teacher responded with this:
P(z>2) = .5-.4772 = .0228 & P(Z<-2) =P(Z>2) by symmetry. This is the area of the upper tail plus that of the lower tail.
So, P(Z<-2)+P(Z>2)= 2P(Z>2) = 2(.0228)
So after obtaining that information, I resubmitted using the above formula:
4.94 A) P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (from the back of my book)= 2(.0228) = .0456
P(Z<-3)+P(Z>3)= 2P(Z>3)
.5 - .4987 (number from the back of my book) = 2(.0013) = .0026
B) P(Z<-1)+P(Z>1)= 2P(Z>1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204
P(Z<-2)+P(Z>2)= 2P(Z>2)
.5 - .4772 (number from the back of my book) = 2(.0228) = .0456
C) P(Z<- .8 )+P(Z> .8 )= 2P(Z>.8 )
.5 - .2281 (number from the back of my book) = .2719 = 2(.2719) = .5438
P(Z<-.1)+P(Z>.1)= 2P(Z>.1)
.5 - .0398 (number from the back of my book) = .4602 = 2(.4602) = .9204
I'm not sure if these even makes sense to anybody but I've been working on this problem for a week now and I have one more week to get it right. Teacher gives us two weeks to make corrections and I am just about lost. I dislike statistics.