Records show that 29% of all payments to a mail-order company are submitted after the due date. Suppose 50 payments are submitted this week. Let r be a random variable that represents the number of payments that are late. Use the normal approximation to the binomial to estimate P(r > 20).
a. 0.0307
b. 0.0594
c. 0.9406
d. 0.9564
Using the information from problem, use the normal approximation to the binomial to estimate P(20< r< 25).
a. 0.0591
b. 0.0585
c. 0.0431
d. 0.0304
0.0594
b) 0.0591
To estimate P(r > 20), we can use the normal approximation to the binomial distribution.
Given:
- The probability of a payment being late is 29% or 0.29.
- The number of payments submitted this week is 50.
We can find the mean (μ) and standard deviation (σ) of the binomial distribution using the following formulas:
μ = n * p = 50 * 0.29 = 14.5
σ = √(n * p * (1 - p)) = √(50 * 0.29 * (1 - 0.29)) ≈ 3.43
To estimate P(r > 20), we can convert the binomial distribution to a normal distribution using the continuity correction and find the z-score:
z = (X - μ) / σ
where X is the value we want to calculate the probability for, which is 20.
z = (20 + 0.5 - μ) / σ
= (20 + 0.5 - 14.5) / 3.43
≈ 1.610
Using the standard normal distribution table or calculator, we can find the probability associated with z = 1.610 to be approximately 0.9472.
Since we want P(r > 20) and not P(r ≥ 20), we need to subtract this value from 1:
P(r > 20) ≈ 1 - 0.9472 = 0.0528
The answer to the first question is a. 0.0307.
To estimate P(20 < r < 25), we can use the same approach of converting the binomial distribution to a normal distribution.
Finding P(20 < r < 25) is equivalent to finding P(r > 20) - P(r > 25).
Using the same z-score formula:
z1 = (20 + 0.5 - μ) / σ
= (20 + 0.5 - 14.5) / 3.43
≈ 1.610
z2 = (25 + 0.5 - μ) / σ
= (25 + 0.5 - 14.5) / 3.43
≈ 3.386
Using the standard normal distribution table or calculator, we can find the probabilities associated with z1 = 1.610 and z2 = 3.386.
P(20 < r < 25) ≈ P(r > 20) - P(r > 25)
≈ 0.9472 - (1 - P(r < 25))
≈ 0.9472 - (1 - P(z < 3.386))
≈ 0.9472 - (1 - 0.9996)
≈ 0.9472 - 0.0004
≈ 0.9468
The answer to the second question is c. 0.0431.
To solve this problem, we need to use the normal approximation to the binomial distribution. The normal approximation is valid when both n * p and n * (1 - p) are greater than or equal to 5, where n is the number of trials and p is the probability of success.
1. For the first part, we want to estimate P(r > 20), where r is the number of payments that are late. We are given that the probability of a payment being late, p, is 0.29. The number of payments this week, n, is 50.
To use the normal approximation, we need to calculate the mean (μ) and standard deviation (σ) of the binomial distribution using the formulas:
μ = n * p
σ = sqrt(n * p * (1 - p))
Substituting the given values:
μ = 50 * 0.29 = 14.5
σ = sqrt(50 * 0.29 * (1 - 0.29)) ≈ 3.6
Now, we want to find P(r > 20), which is equivalent to finding P(r ≥ 21) since r can only take integer values. To use the normal approximation, we convert this to a standardized normal distribution.
z = (x - μ) / σ
Calculating z:
z = (21 - 14.5) / 3.6 ≈ 1.81
Using a standard normal distribution table or calculator, we can find the probability associated with z = 1.81, which is approximately 0.9641.
Now, to find P(r > 20), we subtract this probability from 1:
P(r > 20) ≈ 1 - 0.9641 = 0.0359
Therefore, the answer to the first part of the question is not listed among the options provided. However, the closest option is "a. 0.0307."
2. For the second part, we want to estimate P(20 < r < 25), which is equivalent to finding P(r ≥ 21 and r ≤ 24).
To use the normal approximation, we calculate the probabilities for r = 20 and r = 25, and then subtract the former from the latter.
Calculating z for r = 20:
z1 = (20 - 14.5) / 3.6 ≈ 1.53
Calculating z for r = 25:
z2 = (25 - 14.5) / 3.6 ≈ 2.92
Using a standard normal distribution table or calculator, we can find the probabilities associated with z1 and z2:
P(r ≥ 21) ≈ 1 - probability associated with z1
P(r ≤ 24) ≈ probability associated with z2
Substituting the values, we get:
P(20 < r < 25) ≈ P(r ≥ 21) - P(r ≤ 24)
≈ 1 - probability associated with z1 - probability associated with z2
Now, using the table or calculator, we find the probabilities associated with z1 and z2:
probability associated with z1 ≈ 1 - 0.9382 ≈ 0.0618
probability associated with z2 ≈ 0.9994
Substituting these values:
P(20 < r < 25) ≈ 1 - 0.0618 - 0.9994
≈ 0.9382 - 0.0618
≈ 0.8764
Therefore, the answer to the second part of the question is not listed among the options provided. However, the closest option is "a. 0.0591."