Let X1,X2,…,X25 be random sample ~ normal(0,16) and Y1, Y2,…,Y25 be random sample ~ normal(1,9), assume the two samples are independent ,comput P(X>Y) where X=Sum(Xi)/25 and Y=Sum(Yi)/25
We have two samples, n1=n2=25
both are normally distributed, and are independent.
Under these given conditions,
E(X̄1-X̄2)=E(X̄1)-E(X̄2)=μ1-μ2
VAR(X̄1-X̄2)=V(X̄1)+V(X̄2)=σ1²/n1 + σ2²/n2
Z=(X̄1-X̄2)-(μ1-μ2) / sqrt(σ1²/n1 + σ2²/n2)
should have a N(0,1) distribution.
Thus looking up the normal distribution table will give the required probability if we consider (X̄1-X̄2)=0
Correction to missing parentheses for above expression for Z
Z=[(X̄1-X̄2)-(μ1-μ2)] / sqrt(σ1²/n1 + σ2²/n2)
To compute P(X > Y), we need to find the probability that the sample mean of the X variables is greater than the sample mean of the Y variables.
Given:
X1, X2, ..., X25 ~ N(0, 16)
Y1, Y2, ..., Y25 ~ N(1, 9)
First, let's calculate the means of each sample:
X = (X1 + X2 + ... + X25) / 25
Y = (Y1 + Y2 + ... + Y25) / 25
We can also compute the variances of each sample:
Var(X) = Var(X1 + X2 + ... + X25) / 25 = (Var(X1) + Var(X2) + ... + Var(X25)) / 25 = (16 + 16 + ... + 16) / 25 = 16
Var(Y) = Var(Y1 + Y2 + ... + Y25) / 25 = (Var(Y1) + Var(Y2) + ... + Var(Y25)) / 25 = (9 + 9 + ... + 9) / 25 = 9
Now, we need to calculate the difference of the sample means:
D = X - Y
Since X and Y are independent, the mean of the difference is the difference of the means:
Mean(D) = Mean(X) - Mean(Y) = 0 - 1 = -1
The variance of the difference is the sum of the variances:
Var(D) = Var(X) + Var(Y) = 16 + 9 = 25
The standard deviation is the square root of the variance:
SD(D) = sqrt(Var(D)) = sqrt(25) = 5
To find P(X > Y), we can calculate the Z-score:
Z = (D - Mean(D)) / SD(D)
= (D - (-1)) / 5
= (D + 1) / 5
Now, we can find the probability using the standard normal distribution table or calculator, where we want to find P(Z > (D + 1) / 5).
Please note that we need the specific value of D in order to provide the exact probability.
To compute P(X > Y), we need to find the probability that the average of the X sample is greater than the average of the Y sample.
Let's break down the problem and calculate the means and variances first:
Given that X1, X2, ..., X25 ~ N(0, 16) (mean = 0, variance = 16), the average X = X̄ = (X1 + X2 + ... + X25) / 25.
Similarly, Y1, Y2, ..., Y25 ~ N(1, 9) (mean = 1, variance = 9), the average Y = Ȳ = (Y1 + Y2 + ... + Y25) / 25.
Since X and Y are averages of the random samples, we know that X ~ N(0, 16/25) and Y ~ N(1, 9/25).
Next, we need to find the difference between X and Y. Let Z = X - Y.
Since X and Y are independent random variables, Z follows a normal distribution with mean μZ = μX - μY = 0 - 1 = -1 and variance σ²Z = σ²X + σ²Y = 16/25 + 9/25 = 25/25 = 1.
Now, we can calculate P(Z > 0) using the standard normal distribution table or using statistical software.
Using the standard normal distribution table: P(Z > 0) = 0.5 - P(Z < 0) = 0.5 - P(Z ≤ 0) = 0.5 - 0.5 = 0.
Therefore, P(X > Y) = P(Z > 0) = 0.
Note: The result indicates that the event X > Y is highly unlikely.