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?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related to the Z scores.
To find the probability of the number of vacation days taken by the next employee falling into specific ranges, we can use the concept of the standard normal distribution.
1. Probability that the number of vacation days taken is less than 10 days:
To find this probability, we need to calculate the z-score for the value 10. The z-score is a measure of how many standard deviations a particular value is away from the mean.
The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = the value we want to find the z-score for (10 days)
μ = the mean of the distribution (14 days)
σ = the standard deviation of the distribution (3 days)
Substituting the values into the formula:
z = (10 - 14) / 3 = -1.33
We can then look up the corresponding probability in the z-table or use a calculator. For a z-score of -1.33, the probability is approximately 0.0918, or 9.18%.
Therefore, the probability that the number of vacation days taken by the next employee is less than 10 days is approximately 0.0918 or 9.18%.
2. Probability that the number of vacation days taken is more than 21 days:
Similarly, to find this probability, we need to calculate the z-score for the value 21.
z = (x - μ) / σ
Substituting the values into the formula:
z = (21 - 14) / 3 = 2.33
Looking up the corresponding probability in the z-table or using a calculator, we find that for a z-score of 2.33, the probability is approximately 0.9900, or 99.00%.
Therefore, the probability that the number of vacation days taken by the next employee is more than 21 days is approximately 0.9900, or 99.00%.