An accounting firm is planning for the next tax preparation season. From last year's returns, the firm collects a systematic random sample of 100 filings. The 100 filings showed an average preparation time of 90 minutes with a standard deviation of 140 minutes. What is the probability that the mean completion time is between 1 and 2 hours, i.e., 60 and 120 minutes?
It would help if you proofread your questions before you posted them.
You probably have a typo. With SD = 140, one SD below the mean (90) would be -50. This means that you do not have a normal distribution.
However, if the distribution is normal,
Z = (score-mean)/SD
Insert the correct values to find Z.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to your Z scores.
To find the probability that the mean completion time is between 60 and 120 minutes, we can assume that the distribution of preparation times is approximately normal due to the large sample size (Central Limit Theorem).
First, we need to standardize the values of 60 and 120 minutes using the following formula:
z = (x - μ) / (σ / sqrt(n))
where:
- z is the standard score
- x is the value we want to standardize (60 and 120 minutes in this case)
- μ is the population mean
- σ is the standard deviation
- n is the sample size
In this case, the population mean (μ) is the average preparation time from the sample, which is 90 minutes. The standard deviation (σ) is 140 minutes. The sample size (n) is 100.
Standardizing 60 minutes:
z1 = (60 - 90) / (140 / sqrt(100))
= -30 / (140 / 10)
= -30 / 14
= -2.14
Standardizing 120 minutes:
z2 = (120 - 90) / (140 / sqrt(100))
= 30 / (140 / 10)
= 30 / 14
= 2.14
Now, we need to find the probability that the standard scores fall between -2.14 and 2.14. We can use a standard normal distribution table or calculator to find this probability.
Using the standard normal distribution table, we find that the probability of a standard score being less than -2.14 is approximately 0.0162, and the probability of a standard score being less than 2.14 is also approximately 0.9838.
To find the probability between -2.14 and 2.14, we need to subtract the probability of z being less than -2.14 from the probability of z being less than 2.14:
P(-2.14 < z < 2.14) = P(z < 2.14) - P(z < -2.14)
= 0.9838 - 0.0162
= 0.9676
Therefore, the probability that the mean completion time is between 60 and 120 minutes is approximately 0.9676, or 96.76%.
To find the probability that the mean completion time is between 60 and 120 minutes, we need to use the Central Limit Theorem. According to the Central Limit Theorem, for a large enough sample size, the distribution of sample means will be approximately normal regardless of the shape of the population distribution. Since the sample size is 100, we can assume that the distribution of sample means is approximately normal.
Step 1: Calculate the standard error of the mean (SEM).
The formula to calculate the standard error of the mean is:
SEM = standard deviation / √sample size
In this case, the standard deviation is 140 minutes and the sample size is 100.
SEM = 140 / √100 = 14 minutes
Step 2: Standardize the interval.
To calculate the probability, we need to standardize the interval (60 to 120 minutes) using the formula:
Z = (x - μ) / SEM
Where:
Z is the z-score
x is the desired value (60 or 120 minutes)
μ is the population mean (90 minutes)
SEM is the standard error of the mean (14 minutes)
For x = 60 minutes:
Z = (60 - 90) / 14 = -2.14
For x = 120 minutes:
Z = (120 - 90) / 14 = 2.14
Step 3: Calculate the probability.
Now, we need to find the probability between the two z-scores (-2.14 and 2.14). We can use a standard normal distribution table or a calculator to look up the probabilities corresponding to these z-scores.
Using a standard normal distribution table, we can find that the probability of the mean completion time being between 60 and 120 minutes is approximately 0.8312 or 83.12%.
Therefore, the probability that the mean completion time is between 1 and 2 hours is approximately 0.8312 or 83.12%.