Let X be normally distributed with mean μ = 102 and standard deviation σ = 34. [You may find it useful to reference the z table.]
a. Find P(X ≤ 100). (Round your final answer to 4 decimal places.)
b. Find P(95 ≤ X ≤ 110). (Round your final answer to 4 decimal places.)
c. Find x such that P(X ≤ x) = 0.360. (Round your final answer to 3 decimal places.)
d. Find x such that P(X > x) = 0.830. (Round your final answer to 3 decimal places.)
a. To find P(X ≤ 100), we need to calculate the z-score first. The z-score formula is given by:
z = (x - μ) / σ
Plugging in the values, we have:
z = (100 - 102) / 34 = -0.0588
Now, we can use the z-table to find the probability. Looking up the z-score of -0.0588, we find that the probability is 0.5279.
Therefore, P(X ≤ 100) = 0.5279.
b. To find P(95 ≤ X ≤ 110), we need to calculate the z-scores for both values:
For 95, z = (95 - 102) / 34 = -0.2059
For 110, z = (110 - 102) / 34 = 0.2353
Using the z-table, we can find the probabilities corresponding to these z-scores:
P(X ≤ 95) = 0.4207
P(X ≤ 110) = 0.5899
To find P(95 ≤ X ≤ 110), we subtract the two probabilities:
P(95 ≤ X ≤ 110) = P(X ≤ 110) - P(X ≤ 95) = 0.5899 - 0.4207 = 0.1692
Therefore, P(95 ≤ X ≤ 110) = 0.1692.
c. We are given that P(X ≤ x) = 0.360. To find x, we need to find the z-score corresponding to this probability.
Using the z-table, we find that the z-score for a cumulative probability of 0.360 is approximately -0.3300.
Now, we can use the z-score formula to find x:
-0.3300 = (x - 102) / 34
Solving for x, we have:
-0.3300 * 34 = x - 102
-11.2200 = x - 102
x = -11.2200 + 102
x = 90.7800
Therefore, x ≈ 90.780.
d. We are given that P(X > x) = 0.830. To find x, we need to find the z-score corresponding to this probability.
Since we are looking for the probability greater than x, we subtract the given probability from 1:
P(X > x) = 1 - P(X ≤ x) = 1 - 0.830 = 0.170
Using the z-table, we can find the z-score corresponding to this probability:
The z-score for a cumulative probability of 0.170 is approximately -0.9382.
Now, we can use the z-score formula to find x:
-0.9382 = (x - 102) / 34
Solving for x, we have:
-0.9382 * 34 = x - 102
-31.893 = x - 102
x = -31.893 + 102
x = 70.107
Therefore, x ≈ 70.107.
To solve these problems, we will use the standard normal distribution table, also known as the z-table.
a. Find P(X ≤ 100):
To find P(X ≤ 100), we need to standardize the value 100 using the formula:
z = (x - μ) / σ
where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.
In this case, x = 100, μ = 102, and σ = 34.
z = (100 - 102) / 34 = -0.0588
Now, we can use the z-table to find the probability corresponding to this z-value.
Looking up the z-value -0.0588 in the z-table, we find the corresponding probability to be 0.4750.
Therefore, P(X ≤ 100) = 0.4750 (rounded to 4 decimal places).
b. Find P(95 ≤ X ≤ 110):
To find P(95 ≤ X ≤ 110), we need to standardize the values 95 and 110 using the same formula as before.
z1 = (95 - 102) / 34 = -0.2059
z2 = (110 - 102) / 34 = 0.2353
Now, we can use the z-table to find the probabilities corresponding to these two z-values.
Looking up the z-value -0.2059 in the z-table, we find the corresponding probability to be 0.5822.
Looking up the z-value 0.2353 in the z-table, we find the corresponding probability to be 0.5917.
To find P(95 ≤ X ≤ 110), we subtract the probability of the lower z-value from the probability of the higher z-value:
P(95 ≤ X ≤ 110) = 0.5917 - 0.5822 = 0.0095 (rounded to 4 decimal places).
Therefore, P(95 ≤ X ≤ 110) = 0.0095 (rounded to 4 decimal places).
c. Find x such that P(X ≤ x) = 0.360:
To find x such that P(X ≤ x) = 0.360, we need to use the z-value corresponding to this probability.
Looking up the probability 0.360 in the z-table, we find the corresponding z-value to be -0.3600.
Now, we can use the standardization formula to find x:
-0.3600 = (x - 102) / 34
Solving for x:
-0.3600 * 34 = x - 102
- 12.24 = x - 102
x = 102 - 12.24 = 89.76
Therefore, x ≈ 89.76 (rounded to 3 decimal places).
d. Find x such that P(X > x) = 0.830:
To find x such that P(X > x) = 0.830, we need to find the z-value corresponding to the complement of this probability.
The complement of 0.830 is 1 - 0.830 = 0.170.
Looking up the probability 0.170 in the z-table, we find the corresponding z-value to be -0.954.
Now, we can use the standardization formula to find x:
-0.954 = (x - 102) / 34
Solving for x:
-0.954 * 34 = x - 102
- 32.436 = x - 102
x = 102 - 32.436 = 69.564
Therefore, x ≈ 69.564 (rounded to 3 decimal places).
To find the probabilities and values you're looking for, we can use the standard normal distribution and convert the given values to z-scores. Here's how you can find the answers for each part of the question:
a. Find P(X ≤ 100):
To find this probability, we need to convert it to the corresponding z-score using the formula: z = (X - μ) / σ.
For X = 100, μ = 102, and σ = 34, the z-score is calculated as z = (100 - 102) / 34 = -0.0588 (rounded to 4 decimal places).
Now, you can use a standard normal distribution table (also known as the z-table) to find the probability associated with this z-score. For z = -0.0588, the table will give us a cumulative probability value of 0.4726 (rounded to 4 decimal places).
Therefore, P(X ≤ 100) = 0.4726 (rounded to 4 decimal places).
b. Find P(95 ≤ X ≤ 110):
We need to calculate the z-scores for both X = 95 and X = 110 using the same formula as above. For X = 95, the z-score is (95 - 102) / 34 = -0.2059 (rounded to 4 decimal places), and for X = 110, the z-score is (110 - 102) / 34 = 0.2353 (rounded to 4 decimal places).
Next, we can find the cumulative probabilities associated with both z-scores using the z-table. For z = -0.2059, the table gives us a cumulative probability of 0.5811 (rounded to 4 decimal places), and for z = 0.2353, the table gives us a cumulative probability of 0.5910 (rounded to 4 decimal places).
To find the probability P(95 ≤ X ≤ 110), we subtract the cumulative probability corresponding to the lower z-score from the cumulative probability corresponding to the higher z-score: P(95 ≤ X ≤ 110) = 0.5910 - 0.5811 = 0.0099 (rounded to 4 decimal places).
c. Find x such that P(X ≤ x) = 0.360:
We want to find the value of x (from the original distribution, not z-score) that corresponds to a cumulative probability of 0.360.
To determine the z-score associated with this cumulative probability, we need to check the z-table. The closest cumulative probability to 0.360 in the z-table is 0.3599 at a z-score of approximately -0.3585 (rounded to 4 decimal places).
Now, we can convert this z-score back to the original distribution using the formula: X = z * σ + μ.
For z = -0.3585, σ = 34, and μ = 102, we get x = -0.3585 * 34 + 102 = 89.4829 (rounded to 3 decimal places).
Hence, x such that P(X ≤ x) = 0.360 is approximately 89.483 (rounded to 3 decimal places).
d. Find x such that P(X > x) = 0.830:
In this case, we are asked to find the value of x (from the original distribution) such that the cumulative probability P(X > x) is 0.830.
First, we need to find the z-score associated with a cumulative probability of 0.830.
By referring to the z-table, the closest cumulative probability to 0.830 is 0.8296, which corresponds to a z-score of approximately 0.9864 (rounded to 4 decimal places).
Converting this z-score back to the original distribution using the same formula, X = z * σ + μ, we get x = 0.9864 * 34 + 102 = 134.6156 (rounded to 3 decimal places).
Thus, x such that P(X > x) = 0.830 is approximately 134.616 (rounded to 3 decimal places).
I hope this explanation helps you understand the steps involved in finding the answers!