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?

Standard deviation

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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.

10 days is 1.33 std below the mean

21 days is 2.33 std above the mean

just consult your Z table.

To find the probability that the number of vacation days taken by the next employee is less than 10 days, we need to calculate the z-score and use the standard normal distribution.

Step 1: Calculate the z-score.
The z-score formula is given by: z = (X - μ) / σ, where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case, X = 10, μ = 14, and σ = 3.
So, z = (10 - 14) / 3 = -4 / 3 = -1.33.

Step 2: Look up the z-score in the standard normal distribution table.
By looking up the z-score -1.33 in the standard normal distribution table, we find that the corresponding probability is 0.0918.

Therefore, the probability that the number of days of vacation taken by the next employee is less than 10 days is approximately 0.0918.

To find the probability that the number of days of vacation taken is more than 21 days, we need to calculate the z-score again.

Step 1: Calculate the z-score.
Using the same formula, z = (X - μ) / σ, where X = 21, μ = 14, and σ = 3, we get z = (21 - 14) / 3 = 7 / 3 = 2.33.

Step 2: Calculate the complementary probability.
Since we want to find the probability of more than 21 days, we need to find the corresponding probability of less than 21 days and subtract it from 1.
By looking up the z-score 2.33 in the standard normal distribution table, we find that the corresponding probability is 0.9904.
So, the probability of less than 21 days is 0.9904.
Therefore, the probability of more than 21 days is 1 - 0.9904 = 0.0096.

Therefore, the probability that the number of days of vacation taken by the next employee is more than 21 days is approximately 0.0096.

To find the probabilities, we need to calculate the z-scores for the given values and then find the corresponding area under the standard normal distribution curve.

To find the probability that the number of days of vacation taken is less than 10 days:

Step 1: Calculate the z-score.
Z = (X - μ) / σ
Where:
X = 10 (number of days taken)
μ = 14 (mean)
σ = 3 (standard deviation)

Z = (10 - 14) / 3
Z = -4 / 3

Step 2: Look up the z-score in the Z-table or use a calculator.
The z-table provides the cumulative probability up to a given z-score. In this case, we need the area to the left of the z-score (-4/3), which represents the probability that the number of days taken is less than 10.

Using a z-table or calculator, we find that the area to the left of -4/3 is approximately 0.0918.

Therefore, the probability that the number of days of vacation taken is less than 10 days is approximately 0.0918 or 9.18%.

To find the probability that the number of days of vacation taken is more than 21 days:

Step 1: Calculate the z-score.
Z = (X - μ) / σ
Where:
X = 21 (number of days taken)
μ = 14 (mean)
σ = 3 (standard deviation)

Z = (21 - 14) / 3
Z = 7 / 3

Step 2: Look up the z-score in the Z-table or use a calculator.
The z-table provides the cumulative probability up to a given z-score. In this case, we need the area to the right of the z-score (7/3), which represents the probability that the number of days taken is more than 21.

Using a z-table or calculator, we find that the area to the right of 7/3 is approximately 0.0013.

Therefore, the probability that the number of days of vacation taken is more than 21 days is approximately 0.0013 or 0.13%.