Let Y = ex where X is normally distributed with μ = 2.0 and σ = 0.9. Compute the following values. [You may find it useful to reference the z table.]
a. Compute P(Y ≤ 8.7). (Round your intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
b. Compute P(8.7 < Y < 9.6). (Round your intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)
c. Compute the 60th percentile of Y. (Round your intermediate calculations to at least 4 decimal places and final answer to the nearest whole number.)
To solve these problems, we need to standardize the values and use the standard normal distribution table.
a. To compute P(Y ≤ 8.7), we need to find the z-value corresponding to x = 8.7 first. We can standardize x using the formula:
z = (x - μ) / σ = (8.7 - 2.0) / 0.9 = 7.77
Using the z-table, the probability corresponding to a z-value of 7.77 is very close to 1.0. Therefore, P(Y ≤ 8.7) is approximately 1.0000.
b. To compute P(8.7 < Y < 9.6), we need to find the z-values corresponding to x = 8.7 and x = 9.6.
For x = 8.7:
z1 = (8.7 - 2.0) / 0.9 = 7.77
For x = 9.6:
z2 = (9.6 - 2.0) / 0.9 = 9.33
Using the z-table, the probability corresponding to z1 = 7.77 is very close to 1.0 and the probability corresponding to z2 = 9.33 is very close to 1.0. Therefore, P(8.7 < Y < 9.6) is approximately 1.0000.
c. To compute the 60th percentile of Y, we need to find the corresponding z-value. The 60th percentile means there is a 60% probability that the value of Y is below this percentile.
Using the z-table, we need to find the z-value that corresponds to a cumulative probability of 0.60. The closest value to 0.60 in the table is 0.5793, which corresponds to a z-value of 0.25.
To find the x-value, we can use the formula for standardizing a variable:
z = (x - μ) / σ
Rearranging the formula to solve for x:
x = z * σ + μ
x = 0.25 * 0.9 + 2.0
x ≈ 2.225
Therefore, the 60th percentile of Y is approximately 2.225.
To solve the given problems, we need to use the properties of the normal distribution and the transformation of random variables.
For a random variable Y = ex, where X is normally distributed with mean μ = 2.0 and standard deviation σ = 0.9, we need to find the probabilities and percentiles associated with Y.
a. To compute P(Y ≤ 8.7), we need to first standardize the value of 8.7 using the standardization formula:
z = (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.
In this case, we have:
x = 8.7
μ = 2.0
σ = 0.9
z = (8.7 - 2.0) / 0.9
z = 7.7 / 0.9
z ≈ 8.5556
Next, we need to find the probability associated with the standardized value using the z-table. The probability P(Y ≤ 8.7) is equal to P(Z ≤ 8.5556):
P(Y ≤ 8.7) ≈ P(Z ≤ 8.5556)
Using the z-table, we find that P(Z ≤ 8.5556) ≈ 1.0 (rounded to 4 decimal places).
So, P(Y ≤ 8.7) ≈ 1.0 (rounded to 4 decimal places).
b. To compute P(8.7 < Y < 9.6), we need to standardize the values 8.7 and 9.6 using the formula mentioned above:
For 8.7:
z1 = (8.7 - 2.0) / 0.9
z1 = 7.7 / 0.9
z1 ≈ 8.5556
For 9.6:
z2 = (9.6 - 2.0) / 0.9
z2 = 7.6 / 0.9
z2 ≈ 8.4444
The probability P(8.7 < Y < 9.6) can be obtained by finding the difference between the areas under the standard normal curve corresponding to the two z-values:
P(8.7 < Y < 9.6) = P(Z1 < Z < Z2)
≈ P(8.5556 < Z < 8.4444)
Using the z-table, we can find P(8.5556 < Z < 8.4444) ≈ 0 (rounded to 4 decimal places).
So, P(8.7 < Y < 9.6) ≈ 0 (rounded to 4 decimal places).
c. To compute the 60th percentile of Y, we need to find the value of Y that corresponds to the cumulative probability of 0.6.
To do this, we need to find the z-value associated with the cumulative probability of 0.6 using the z-table. The corresponding z-value is approximately 0.2533 (rounded to 4 decimal places).
Next, we can use the standardization formula mentioned above to find the value of Y:
z = (x - μ) / σ
For the given problem, we have:
x = Y
μ = 2.0
σ = 0.9
z = 0.2533
Plugging these values into the formula, we have:
0.2533 = (Y - 2.0) / 0.9
Solving for Y:
0.2533 * 0.9 = Y - 2.0
0.228 = Y - 2.0
Y ≈ 2.228
So, the 60th percentile of Y is approximately 2 (rounded to the nearest whole number).
To compute the values in the given problem, we first need to standardize the random variable X using the formula:
Z = (X - μ) / σ
a. Compute P(Y ≤ 8.7):
To find this probability, we need to convert 8.7 into a standardized value (Z-score) using the formula mentioned above.
Z = (8.7 - 2.0) / 0.9 = 8.0
Now, we need to find the cumulative probability associated with Z = 8.0 from the standard normal distribution table.
Using the z-table, we find that the cumulative probability for Z = 8.0 is approximately 1.
Therefore, P(Y ≤ 8.7) = 1.
b. Compute P(8.7 < Y < 9.6):
Similarly, we need to standardize both 8.7 and 9.6 to find the corresponding Z-scores.
Z1 = (8.7 - 2.0) / 0.9 = 8.0
Z2 = (9.6 - 2.0) / 0.9 = 9.0
Now, we need to find the difference between the two Z-scores, which represents the probability between the two values.
P(8.7 < Y < 9.6) = P(Z1 < Z < Z2)
Using the z-table, we find the cumulative probabilities for Z1 = 8.0 and Z2 = 9.0:
P(Z < 8.0) = 1 (from a.)
P(Z < 9.0) ≈ 1 (from the z-table)
Thus, P(8.7 < Y < 9.6) ≈ 1 - 1 = 0.
c. Compute the 60th percentile of Y:
To find the 60th percentile of Y, we need to find the Z-score corresponding to the cumulative probability of 0.60 from the standard normal distribution table.
Using the z-table, we find the Z-score associated with the 60th percentile to be approximately 0.253.
Now, we can use the Z-score formula to find the corresponding value of Y:
Z = (X - μ) / σ
0.253 = (X - 2.0) / 0.9
Solving for X:
X - 2.0 = 0.253 * 0.9
X ≈ 2.227
Rounding this value to the nearest whole number, we get the 60th percentile of Y to be approximately 2.