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.
P(X>0.75)=
P(X≤−1.25)=
Let Z=(Y−3)/4. Find the mean and the variance of Z.
E[Z]=
var(Z)=
P(−1≤Y≤2)=
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
Oh, statistics, the study of never-ending probabilities! Strap on your mathematical boots and let's dive into your questions!
First, let's tackle P(X>0.75). Now, we know that X is a normal random variable with mean 0 and variance 1. To find this probability, we need to convert it to a standard normal distribution using the formula Z = (X - μ) / σ, where Z is the standard normal random variable with mean 0 and variance 1.
So, for P(X > 0.75), we need to find P(Z > (0.75 - 0) / sqrt(1)). Let's use the standard normal table for that.
*looking at the table, scratching head*
Ah, here it is! P(Z > 0.75) = 1 - P(Z <= 0.75). According to the table, P(Z <= 0.75) is 0.7734. So, P(X > 0.75) = 1 - 0.7734 = 0.2266.
Next up, P(X <= -1.25). Again, we need to convert X to a standard normal random variable first. Using the same formula Z = (X - μ) / σ, we find Z = (-1.25 - 0) / sqrt(1) = -1.25. Looking at the standard normal table, we find P(Z <= -1.25) = 0.1056. So, P(X <= -1.25) = 0.1056.
Now let's move on to Z = (Y - 3) / 4. We want to find the mean and variance of Z. Well, knowing clown logic, the mean of Z is just the mean of Y, which is 2. So, E[Z] = 2.
For the variance, we use the properties of linear transformations. Remember that var(aX + b) = a^2 * var(X). In our case, var(Z) = var((Y - 3) / 4) = (1/4)^2 * var(Y) = 1/16 * 9 = 0.5625.
Finally, let's take a stab at P(-1 ≤ Y ≤ 2). Since Y is already normal with mean 2 and variance 9, we don't need to convert it. We can use the standard normal table, looking for P(-1 ≤ Y ≤ 2). Unfortunately, the standard normal table only gives probabilities for the standard normal distribution with mean 0 and variance 1. So, we'll need to standardize our values first.
For -1, we have Z1 = (-1 - 2) / 3 = -1. Now, for 2, we have Z2 = (2 - 2) / 3 = 0.
Applying the standard normal table for -1 and 0, we get P(Z <= -1) = 0.1587 and P(Z <= 0) = 0.5000. Subtracting these two probabilities gives us P(-1 ≤ Y ≤ 2) = 0.5000 - 0.1587 = 0.3413.
There you have it – my probabilistic circus act is complete! I hope this both amused and helped you with your questions.
To find the values using the standard normal table, we'll need to standardize the variables using z-scores.
1. Find P(X > 0.75):
To standardize X, we use the formula: z = (x - μ) / σ, where x is the value of X, μ is the mean, and σ is the standard deviation.
In this case, X has a mean of 0 and a standard deviation of 1.
So, z = (0.75 - 0) / 1 = 0.75.
Looking up the z-score of 0.75 in the standard normal table, we find the corresponding probability to be 0.7734.
Therefore, P(X > 0.75) = 1 - P(X ≤ 0.75) = 1 - 0.7734 = 0.2266.
2. Find P(X ≤ -1.25):
Standardizing X, we get z = (-1.25 - 0) / 1 = -1.25.
Looking up the z-score of -1.25 in the standard normal table, we find the corresponding probability to be 0.1056.
Therefore, P(X ≤ -1.25) = 0.1056.
3. Let Z = (Y - 3) / 4.
To find the mean and variance of Z, we can use the properties of linear transformations.
The mean of Z is given by E[Z] = (mean of Y - 3) / 4 = (2 - 3) / 4 = -1 / 4 = -0.25.
To find the variance of Z, we use the property that Var(aX + b) = a^2 Var(X) for any constants a and b. In this case, a = 1/4.
Var(Z) = Var((Y - 3) / 4) = (1/4)^2 Var(Y) = (1/4)^2 * 9 = 9/16 = 0.5625.
4. Find P(-1 ≤ Y ≤ 2):
Standardizing the values, we get z1 = (-1 - 2) / 3 = -1 and z2 = (2 - 2) / 3 = 0.
Using the standard normal table, we find P(-1 ≤ Z ≤ 0) = 0.3413.
Therefore, P(-1 ≤ Y ≤ 2) = P(-1 ≤ Z ≤ 0) = 0.3413.
In summary:
P(X > 0.75) ≈ 0.2266
P(X ≤ -1.25) ≈ 0.1056
E[Z] = -0.25
Var(Z) = 0.5625
P(-1 ≤ Y ≤ 2) = 0.3413.
To find the probabilities and expected values using the standard normal table, we need to standardize the random variables by converting them into standard normal variables.
1. P(X > 0.75):
First, we calculate the standard score (Z-score) for X = 0.75 using the formula:
Z = (X - mean) / standard deviation
For X:
Mean = 0
Standard deviation = 1
Z = (0.75 - 0) / 1 = 0.75
Now, we look up the probability P(Z > 0.75) in the standard normal table. The standard normal table provides the cumulative probability up to a certain Z-value.
From the table, we find that P(Z > 0.75) = 1 - P(Z ≤ 0.75)
Using the standard normal table, we find:
P(Z ≤ 0.75) = 0.7734
Therefore, P(X > 0.75) = 1 - P(Z ≤ 0.75) = 1 - 0.7734 = 0.2266
2. P(X ≤ -1.25):
Similarly, we calculate the standard score for X = -1.25:
Z = (X - mean) / standard deviation
For X:
Mean = 0
Standard deviation = 1
Z = (-1.25 - 0) / 1 = -1.25
From the standard normal table, we find P(Z ≤ -1.25) = 0.1056
Therefore, P(X ≤ -1.25) = P(Z ≤ -1.25) = 0.1056
3. Let Z = (Y - 3) / 4. We need to find the mean and variance of Z.
For Z, the mean and variance can be calculated as follows:
E[Z] = E[(Y - 3) / 4]
= (1/4) * (E[Y] - 3)
For Y:
Mean = 2
Standard deviation = 3 (the square root of variance)
E[Z] = (1/4) * (2 - 3) = -1/4 = -0.25
var(Z) = var((Y - 3) / 4)
= (1/16) * var(Y)
For Y:
Variance = 9
var(Z) = (1/16) * 9 = 0.5625
4. P(-1 ≤ Y ≤ 2):
To find this probability, we standardize the given values and then calculate the cumulative probability using the standard normal table.
For -1:
Z = (X - mean) / standard deviation
For Y:
Mean = 2
Standard deviation = 3 (the square root of variance)
Z1 = (-1 - 2) / 3 = -3/3 = -1
For 2:
Z2 = (2 - 2) / 3 = 0/3 = 0
To find P(-1 ≤ Y ≤ 2), we need to calculate P(-1 ≤ Z ≤ 0) using the standard normal table.
From the table, we find P(-1 ≤ Z ≤ 0) = P(Z ≤ 0) - P(Z ≤ -1)
P(Z ≤ 0) = 0.5 (from the standard normal table)
P(Z ≤ -1) = 0.1587 (from the standard normal table)
P(-1 ≤ Y ≤ 2) = P(-1 ≤ Z ≤ 0) = 0.5 - 0.1587 = 0.3413
So, P(-1 ≤ Y ≤ 2) = 0.3413