the average monthly mortgage payment for all homeowners in a city is 2760. suppose that the distribution of monthly mortgages paid by homeowners in this city follow an approximately normal distribution with a mean of 2760 and a standard deviation of 460. find to 4 decimal places that the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800
To find the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800, we need to standardize the values using the z-score formula and then find the area under the normal curve.
First, we determine the z-score for the lower value of 3300:
z1 = (3300 - 2760) / 460 = 1.1739
Next, we determine the z-score for the upper value of 3800:
z2 = (3800 - 2760) / 460 = 2.2609
Using a standard normal distribution table or a calculator, we find the cumulative probability associated with each z-score:
P(z1 < Z < z2) = P(1.1739 < Z < 2.2609) = 0.8647 – 0.8790 = -0.0143
Since the area under the curve cannot be negative, we take the absolute value of -0.0143:
P(3300 < X < 3800) = | -0.0143 | = 0.0143
Therefore, the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800 is approximately 0.0143.
To find the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800, you will need to standardize the values and use the standard normal distribution.
Step 1: Standardize the values
To standardize the values, we need to convert them into z-scores. The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = given value
μ = mean
σ = standard deviation
For the lower bound (3300):
z1 = (3300 - 2760) / 460
For the upper bound (3800):
z2 = (3800 - 2760) / 460
Step 2: Find the probabilities associated with the z-scores
Using a standard normal distribution table or a calculator, we can find the probabilities associated with the z-scores calculated above.
Using a standard normal distribution table, the probability corresponding to the lower bound (z1) is P(Z < z1) and the probability corresponding to the upper bound (z2) is P(Z < z2).
Step 3: Calculate the final probability
To find the probability that the monthly mortgage paid by a randomly selected homeowner is between 3300 and 3800, we subtract the cumulative probability of the lower bound from the cumulative probability of the upper bound.
P(3300 < X < 3800) = P(Z < z2) - P(Z < z1)
Substituting the z-scores back:
P(3300 < X < 3800) = P(Z < (3800 - 2760) / 460) - P(Z < (3300 - 2760) / 460)
Now, let's calculate the probabilities:
z1 = (3300 - 2760) / 460 = 1.1522
z2 = (3800 - 2760) / 460 = 2.2609
Using a standard normal distribution table or a calculator, we find that:
P(Z < z1) = 0.8749
P(Z < z2) = 0.9871
Therefore, the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800 is:
P(3300 < X < 3800) = 0.9871 - 0.8749 = 0.1122
Rounded to 4 decimal places, the probability is approximately 0.1122.
To find the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800, we need to use the concept of the standard normal distribution.
Step 1: Standardize the values
We start by standardizing the values of 3300 and 3800 using the formula:
z = (x - μ) / σ
Where:
x is the value we want to standardize,
μ is the mean of the distribution, and
σ is the standard deviation of the distribution.
For 3300:
z1 = (3300 - 2760) / 460
For 3800:
z2 = (3800 - 2760) / 460
Step 2: Look up the probabilities
Next, we need to find the probabilities associated with these standardized values. We can refer to a standard normal distribution table, or use a calculator with a built-in function to find the corresponding probabilities.
Using a calculator or a standard normal distribution table, we find the following probabilities:
P(z1 < Z < z2) = P(z1 < Z < z2) ≈ P(-0.7391 < Z < 1.5435)
Step 3: Find the probability
Now, subtract the value of the cumulative standard normal distribution corresponding to z1 from the value corresponding to z2.
P(z1 < Z < z2) = P(-0.7391 < Z < 1.5435)
≈ P(Z < 1.5435) - P(Z < -0.7391)
Using the standard normal distribution table or a calculator, we find:
P(Z < 1.5435) ≈ 0.9382
P(Z < -0.7391) ≈ 0.2290
P(z1 < Z < z2) ≈ 0.9382 - 0.2290
≈ 0.7092 (rounded to 4 decimal places)
Therefore, the probability that the monthly mortgage paid by a randomly selected homeowner from this city is between 3300 and 3800 is approximately 0.7092.