10 days is 1.33 std below the mean
21 days is 2.33 std above the mean
just consult your Z table.
The number of vacation days taken by employees of a company is normally distributed with a mean of 14 days and a standard deviation of 3 days. For the next employee, what is the probability that the number of days of vacation taken is less than 10 days? More than 21 days?
2 answers
Standard deviation
Student’s Name:
Institutional affiliation:
Date:
For the normal distribution, Pr(X<=x) =x-µ/δ in which x is the standard data point, µ represents the mean while δ is the standard deviation of the data provided.
With n=10, mean=14 and standard deviation=3,
P (x ≤ 10), z = (10-14) /3 = -4/3,
P (z ≤ -1.3333) = 9.12%
With n=21, with mean still 14 and standard deviation as 3,
P (x ≥ 21), z = (21-14)/3 = 7/3,
P (x ≥ 21) = 1 - P (x <21) = 1 - P (z < 2.3333) = .0098 or. 98%
We compare these answers to what is obtained using the normal distribution online calculator.
When using a mean of 14 and a standard deviation of 3 days, the probabilities that the number of days employees will be on vacation for less than 10 days obtained generates almost similar responses. However, deviation from the mean is higher for the population than in the sample. I.e. 11-17 days for the sample compared to 5-23 days for the population. The normal distribution online calculator generates results while taking the data as a population. The difference in variance between the two can be attributed to the fact that population standard deviation puts into consideration every data point. This ensures that there are less errors as compared to when using the sample of the population. It is thus highly recommendable to use the entire units if in need of more correct responses.
Student’s Name:
Institutional affiliation:
Date:
For the normal distribution, Pr(X<=x) =x-µ/δ in which x is the standard data point, µ represents the mean while δ is the standard deviation of the data provided.
With n=10, mean=14 and standard deviation=3,
P (x ≤ 10), z = (10-14) /3 = -4/3,
P (z ≤ -1.3333) = 9.12%
With n=21, with mean still 14 and standard deviation as 3,
P (x ≥ 21), z = (21-14)/3 = 7/3,
P (x ≥ 21) = 1 - P (x <21) = 1 - P (z < 2.3333) = .0098 or. 98%
We compare these answers to what is obtained using the normal distribution online calculator.
When using a mean of 14 and a standard deviation of 3 days, the probabilities that the number of days employees will be on vacation for less than 10 days obtained generates almost similar responses. However, deviation from the mean is higher for the population than in the sample. I.e. 11-17 days for the sample compared to 5-23 days for the population. The normal distribution online calculator generates results while taking the data as a population. The difference in variance between the two can be attributed to the fact that population standard deviation puts into consideration every data point. This ensures that there are less errors as compared to when using the sample of the population. It is thus highly recommendable to use the entire units if in need of more correct responses.
1 answer