Suppose the yearly rainfall totals for a some city follow a normal distribution, with mean of 18 inches and standard deviation of 6 inches. For a randomly selected year, what is the probability, P, that total rainfall will be in each of the following intervals? (Round all answers to four decimal places.)
(a) Less than 12 inches.
P = ?
(b) Greater than 27 inches.
P = ?
(c) Between 12 and 24 inches.
P = ?
(d) Greater than 35 inches.
P = ?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores you find.
To solve this problem, we will use the standard normal distribution table.
(a) To find the probability of rainfall being less than 12 inches, we need to calculate the z-score first.
z = (x - μ) / σ
where x is the value of rainfall, μ is the mean, and σ is the standard deviation.
For x = 12, μ = 18, σ = 6:
z = (12 - 18) / 6 = -1
Now, using the standard normal distribution table, we find the probability associated with a z-score of -1.
The table shows that the value for z = -1 is 0.1587.
Therefore, P(rainfall < 12 inches) = 0.1587.
(b) To find the probability of rainfall being greater than 27 inches, we again need to calculate the z-score.
For x = 27, μ = 18, σ = 6:
z = (27 - 18) / 6 = 1.5
Using the standard normal distribution table, we find the probability associated with a z-score of 1.5.
The table shows that the value for z = 1.5 is 0.9332.
However, we need to find the probability of rainfall being greater than 27 inches, so we subtract this value from 1.
P(rainfall > 27 inches) = 1 - 0.9332 = 0.0668.
(c) To find the probability of rainfall being between 12 and 24 inches, we calculate the z-scores for both values.
For x = 12, μ = 18, σ = 6:
z1 = (12 - 18) / 6 = -1
For x = 24, μ = 18, σ = 6:
z2 = (24 - 18) / 6 = 1
We need to find the area between z = -1 and z = 1 on the standard normal distribution curve.
Using the standard normal distribution table, we find the probability associated with z = -1 as 0.1587 and with z = 1 as 0.8413.
The area between these two z-values is given by subtracting the smaller probability from the larger probability:
P(12 < rainfall < 24 inches) = 0.8413 - 0.1587 = 0.6826.
(d) To find the probability of rainfall being greater than 35 inches, we calculate the z-score.
For x = 35, μ = 18, σ = 6:
z = (35 - 18) / 6 = 2.8333
Using the standard normal distribution table, we find the probability associated with z = 2.8333 as 0.9977.
However, we need to find the probability of rainfall being greater than 35 inches, so we subtract this value from 1.
P(rainfall > 35 inches) = 1 - 0.9977 = 0.0023
To find the probability in each interval, we need to use the standard normal distribution and the given mean and standard deviation.
(a) Finding the probability of less than 12 inches:
First, we need to convert the rainfall value of 12 inches to a z-score. The z-score formula is given by:
z = (x - mean) / standard deviation
Plugging in the values, we get:
z = (12 - 18) / 6
z = -1
Now, we can use a standard normal distribution table or a calculator to find the probability corresponding to a z-score of -1. The table or calculator will give us the area under the standard normal curve to the left of -1.
Using a table or calculator, we find that the probability of a z-score less than -1 is approximately 0.1587. Therefore, the probability of having less than 12 inches of rainfall is 0.1587.
(b) Finding the probability of greater than 27 inches:
Similarly, we need to convert the rainfall value of 27 inches to a z-score:
z = (27 - 18) / 6
z = 1.5
Using the standard normal distribution table or a calculator, we find that the probability of a z-score greater than 1.5 is approximately 0.0668.
However, we are interested in the probability of greater than 27 inches, so we need to find the area to the right of 1.5. This area can be found by subtracting the probability from 1:
P(greater than 27 inches) = 1 - 0.0668
P(greater than 27 inches) = 0.9332
Therefore, the probability of having more than 27 inches of rainfall is 0.9332.
(c) Finding the probability between 12 and 24 inches:
We need to find the probability of the rainfall being between 12 and 24 inches. Since the distribution is symmetrical, we can find this probability by calculating the area under the standard normal curve to the left of 24 inches and subtracting the area to the left of 12 inches.
Converting both values to z-scores, we get:
z1 = (12 - 18) / 6
z1 = -1
z2 = (24 - 18) / 6
z2 = 1
Using the standard normal distribution table or a calculator, we find the probability of a z-score less than -1 is 0.1587, and the probability of a z-score less than 1 is 0.8413.
Therefore, the probability of having rainfall between 12 and 24 inches is:
P(between 12 and 24 inches) = 0.8413 - 0.1587
P(between 12 and 24 inches) = 0.6826
Hence, the probability of having rainfall between 12 and 24 inches is 0.6826.
(d) Finding the probability of greater than 35 inches:
We first convert the rainfall value of 35 inches to a z-score:
z = (35 - 18) / 6
z = 2.8333
Using the standard normal distribution table or a calculator, we find that the probability of a z-score greater than 2.8333 is approximately 0.0023.
Since we are interested in the probability of greater than 35 inches, we subtract this value from 1:
P(greater than 35 inches) = 1 - 0.0023
P(greater than 35 inches) = 0.9977
Therefore, the probability of having more than 35 inches of rainfall is 0.9977.