Let X be normally distributed with mean μ = 26 and standard deviation σ = 13. [You may find it useful to reference the z table.]
a. Find P(X ≤ 0). (Round your final answer to 4 decimal places.)
b. Find P(X > 13). (Round your final answer to 4 decimal places.)
c. Find P(13 ≤ X ≤ 26). (Round your final answer to 4 decimal places.)
d. Find P(26 ≤ X ≤ 39). (Round your final answer to 4 decimal places.)
To solve this problem, we can use the z-score formula, which is given as:
z = (X - μ) / σ
where X is the random variable, μ is the mean, and σ is the standard deviation.
a. To find P(X ≤ 0), we need to find the z-score corresponding to 0 and then look up the corresponding probability in the z-table.
z = (0 - 26) / 13
z = -26 / 13
z = -2
Looking up the z-score of -2 in the z-table, we find that the corresponding probability is 0.0228.
Therefore, P(X ≤ 0) = 0.0228.
b. To find P(X > 13), we can find the probability of X ≤ 13 and subtract it from 1.
z = (13 - 26) / 13
z = -13 / 13
z = -1
Looking up the z-score of -1 in the z-table, we find that the corresponding probability is 0.1587.
P(X ≤ 13) = 0.1587
P(X > 13) = 1 - P(X ≤ 13) = 1 - 0.1587 = 0.8413
Therefore, P(X > 13) = 0.8413.
c. To find P(13 ≤ X ≤ 26), we can find the probability of X ≤ 26 and subtract the probability of X ≤ 13.
P(X ≤ 13) = 0.1587 (from previous calculations)
P(X ≤ 26) = 0.5 (since it is half of the distribution)
P(13 ≤ X ≤ 26) = P(X ≤ 26) - P(X ≤ 13) = 0.5 - 0.1587 = 0.3413
Therefore, P(13 ≤ X ≤ 26) = 0.3413.
d. To find P(26 ≤ X ≤ 39), we can find the probability of X ≤ 39 and subtract the probability of X ≤ 26.
P(X ≤ 26) = 0.5 (since it is half of the distribution)
P(X ≤ 39) = 0.8413 (from previous calculations)
P(26 ≤ X ≤ 39) = P(X ≤ 39) - P(X ≤ 26) = 0.8413 - 0.5 = 0.3413
Therefore, P(26 ≤ X ≤ 39) = 0.3413.
To solve these questions, we need to calculate the z-scores and use the z-table.
a. To find P(X ≤ 0), we need to calculate the z-score for 0 using the formula:
z = (x - μ) / σ.
Plugging in the values:
z = (0 - 26) / 13 = -26 / 13 = -2.
Using the z-table, the probability corresponding to a z-score of -2 is 0.0228.
Therefore, P(X ≤ 0) = 0.0228.
b. To find P(X > 13), we need to calculate the z-score for 13 using the formula:
z = (x - μ) / σ.
Plugging in the values:
z = (13 - 26) / 13 = -13 / 13 = -1.
Using the z-table, the probability corresponding to a z-score of -1 is 0.1587.
Since we want to find the probability that X is greater than 13, we subtract this probability from 1:
P(X > 13) = 1 - 0.1587 = 0.8413.
c. To find P(13 ≤ X ≤ 26), we need to calculate the z-scores for 13 and 26 using the formula:
z = (x - μ) / σ.
For 13:
z = (13 - 26) / 13 = -13 / 13 = -1.
For 26:
z = (26 - 26) / 13 = 0.
Using the z-table, the probability corresponding to a z-score of -1 is 0.1587, and the probability corresponding to a z-score of 0 is 0.5.
To find the probability between 13 and 26, we subtract the probability corresponding to the z-score of -1 from the probability corresponding to the z-score of 0:
P(13 ≤ X ≤ 26) = 0.5 - 0.1587 = 0.3413.
d. To find P(26 ≤ X ≤ 39), we need to calculate the z-scores for 26 and 39 using the formula:
z = (x - μ) / σ.
For 26:
z = (26 - 26) / 13 = 0.
For 39:
z = (39 - 26) / 13 = 13 / 13 = 1.
Using the z-table, the probability corresponding to a z-score of 0 is 0.5, and the probability corresponding to a z-score of 1 is 0.8413.
To find the probability between 26 and 39, we subtract the probability corresponding to the z-score of 0 from the probability corresponding to the z-score of 1:
P(26 ≤ X ≤ 39) = 0.8413 - 0.5 = 0.3413.
To find the probabilities in this normal distribution problem, we'll need to standardize the values using the standard normal distribution or the z-table.
The z-score, denoted as z, is calculated using the formula:
z = (X - μ) / σ
where X is the value we're interested in, μ is the mean, and σ is the standard deviation.
a. To find P(X ≤ 0), we need to find the probability to the left of 0. First, we calculate the z-score:
z = (0 - 26) / 13
z = -2
Looking for the z-value of -2 in the z-table, we find that the corresponding probability is 0.0228. Thus, P(X ≤ 0) = 0.0228.
b. To find P(X > 13), we need to find the probability to the right of 13. Again, we calculate the z-score:
z = (13 - 26) / 13
z = -1
Looking for the z-value of -1 in the z-table, we find that the corresponding probability is 0.1587. However, we want the probability to the right of 13, so we subtract 0.1587 from 1:
P(X > 13) = 1 - 0.1587 = 0.8413.
c. To find P(13 ≤ X ≤ 26), we need to find the probability between 13 and 26. We calculate the z-scores:
z1 = (13 - 26) / 13 = -1
z2 = (26 - 26) / 13 = 0
Looking for the z-values of -1 and 0 in the z-table, we find that the probabilities are 0.1587 and 0.5000 respectively.
To find P(13 ≤ X ≤ 26), we subtract the probability of being less than 13 from the probability of being less than or equal to 26:
P(13 ≤ X ≤ 26) = 0.5000 - 0.1587 = 0.3413.
d. To find P(26 ≤ X ≤ 39), we follow a similar process. We calculate the z-scores:
z1 = (26 - 26) / 13 = 0
z2 = (39 - 26) / 13 = 1
Looking for the z-values of 0 and 1 in the z-table, we find that the probabilities are 0.5000 and 0.8413 respectively.
To find P(26 ≤ X ≤ 39), we subtract the probability of being less than 26 from the probability of being less than or equal to 39:
P(26 ≤ X ≤ 39) = 0.8413 - 0.5000 = 0.3413.
So, P(X ≤ 0) = 0.0228, P(X > 13) = 0.8413, P(13 ≤ X ≤ 26) = 0.3413, and P(26 ≤ X ≤ 39) = 0.3413.