A firm claims that only 10% of it's accounts receivables are over 30 days past due. The bank that supplies working capital or interim loans to the firms is suspect of the figure and accordingly takes a random sample of 100 accounts of the firm. Find the probability that the sample proportionality be:
a)between 9% and 10%.
b)at least 12%
To find the probability in both cases, we can use the concept of sampling distributions and the normal distribution.
A) To find the probability of the sample proportion being between 9% and 10%, we need to find the area under the normal curve between these two values.
1. Determine the mean of the sample proportion: The mean of the sample proportion is equal to the claimed proportion, which is 10% or 0.1.
2. Determine the standard deviation of the sample proportion: To find the standard deviation, we can use the formula sqrt(p(1-p)/n), where p is the claimed proportion (0.1) and n is the sample size (100).
Standard deviation (σ) = sqrt(0.1 * (1 - 0.1) / 100) ≈ 0.03
3. Convert the sample proportion values into z-scores: To use the normal distribution table, we need to convert the values into standardized z-scores. The formula for z-score is z = (x - μ) / σ, where x is the sample proportion value, μ is the mean, and σ is the standard deviation.
For 9%: z = (0.09 - 0.1) / 0.03 ≈ -0.33
For 10%: z = (0.1 - 0.1) / 0.03 = 0
4. Find the probability using the normal distribution table: Using the z-scores, we can find the corresponding probabilities from the normal distribution table.
P(9% ≤ x ≤ 10%) = P(-0.33 ≤ z ≤ 0) = P(z ≤ 0) - P(z ≤ -0.33)
Looking up the values in the normal distribution table, we find that P(z ≤ 0) is approximately 0.5 and P(z ≤ -0.33) is approximately 0.3707.
P(9% ≤ x ≤ 10%) ≈ 0.5 - 0.3707 = 0.1293
Therefore, the probability that the sample proportion is between 9% and 10% is approximately 0.1293.
B) To find the probability of the sample proportion being at least 12%, we need to find the area under the normal curve from 12% to 100%.
1. Convert the sample proportion value into a z-score: Using the formula z = (x - μ) / σ, where x is the sample proportion value (0.12), μ is the mean (0.1), and σ is the standard deviation (0.03).
z = (0.12 - 0.1) / 0.03 ≈ 0.67
2. Find the probability using the normal distribution table: Using the z-score, we can find the corresponding probability from the normal distribution table.
P(x ≥ 12%) = 1 - P(x ≤ 0.67)
Looking up the value in the normal distribution table, we find that P(z ≤ 0.67) is approximately 0.7486.
P(x ≥ 12%) ≈ 1 - 0.7486 = 0.2514
Therefore, the probability that the sample proportion is at least 12% is approximately 0.2514.