Let X and Y be normal random variables with means 0 and 2, respectively, and variances 1 and 9, respectively. Find the following, using the standard normal table. Express your answers to an accuracy of 4 decimal places.
1 P(X>0.75)=
2 P(X≤−1.25)=
3 Let Z=(Y−3)/4. Find the mean and the variance of Z.
4 E[Z]=
var(Z)=
5 P(−1≤Y≤2)=
anyone know the answers to this?
1 P(X>0.75)= 0.2266
2 P(X≤−1.25)= 0.1056
Let Z=(Y−3)/4. Find the mean and the variance of Z.
3. E[Z]= -0.25
4. var(Z)= 0.5625
5. P(−1≤Y≤2)= 0.3413
1 To find P(X>0.75), we need to calculate the standard normal value that corresponds to 0.75 and then subtract that value from 1.
Using the standard normal table, we find that the standard normal value corresponding to 0.75 is 0.7734. Therefore, P(X>0.75) = 1 - 0.7734 = 0.2266.
2 To find P(X ≤ -1.25), we need to calculate the standard normal value that corresponds to -1.25.
Using the standard normal table, we find that the standard normal value corresponding to -1.25 is 0.1056. Therefore, P(X ≤ -1.25) = 0.1056.
3 To find the mean and variance of Z, we need to substitute the given values and calculate the resulting values.
Given Z = (Y - 3) / 4, we know that the mean of Z is equal to the mean of Y minus 3, divided by 4.
Mean of Z = (mean of Y - 3) / 4 = (2 - 3) / 4 = -1/4.
The variance of Z is equal to the variance of Y divided by 4 squared.
Variance of Z = Variance of Y / (4 squared) = 9 / 16 = 0.5625.
4 E[Z] is the mean of Z, which we have already calculated as -1/4.
E[Z] = -1/4.
5 To find P(-1 ≤ Y ≤ 2), we need to calculate the standard normal values that correspond to -1 and 2, and then subtract the smaller value from the larger value.
Using the standard normal table, we find that the standard normal value corresponding to -1 is 0.1587, and the standard normal value corresponding to 2 is 0.9772.
P(-1 ≤ Y ≤ 2) = 0.9772 - 0.1587 = 0.8185.
To find the following probabilities and values using the standard normal table, we need to standardize the normal random variables X and Y.
1. P(X > 0.75):
First, we standardize X by subtracting its mean and dividing by its standard deviation:
Z_X = (0.75 - 0) / 1 = 0.75
Looking at the standard normal table, we find the probability corresponding to Z_X = 0.75 as P(Z > 0.75) = 1 - P(Z ≤ 0.75).
2. P(X ≤ -1.25):
We standardize X by subtracting its mean and dividing by its standard deviation:
Z_X = (-1.25 - 0) / 1 = -1.25
Looking at the standard normal table, we find the probability corresponding to Z_X = -1.25 as P(Z ≤ -1.25).
3. To find the mean and variance of the random variable Z, we substitute the values from Y into the formula:
Z = (Y - 3) / 4
Mean of Z:
E[Z] = (E[Y] - 3) / 4
Since Y has a mean of 2:
E[Z] = (2 - 3) / 4
Variance of Z:
var(Z) = (var(Y)) / (4^2)
Since Y has a variance of 9:
var(Z) = 9 / (4^2)
4. E[Z]:
Substitute the mean of Y into the formula for Z:
E[Z] = (2 - 3) / 4
5. P(-1 ≤ Y ≤ 2):
We standardize both endpoints of the range:
Z_lower = (-1 - 2) / 3
Z_upper = (2 - 2) / 3
Therefore, we need to find P(Z_lower ≤ Z ≤ Z_upper) using the standard normal table.
Using these steps and the provided formulas, you can now solve for the required probabilities and values. Remember to consult the standard normal table to find the corresponding probabilities for the standardized values.