The number of vacation days taken by the employees of a company is normally distributed with a mean of 14 days and a standard deviation of 3 days. Is this a case of sample standard deviation or population standard deviation? What are some differences between sample standard deviation and population standard deviation? For the next employee, what is the probability that the number of days of vacation taken is less than 10 days? What is the probability that the number of days of vacation taken is more than 21 days? Discuss the solutions and an explanation. Respond to your peers' posts and offer your feedback on this topic.
Standard deviation is a measure of the spread of data distribution (Laerd Statistics, n.d.) . The degree of standard deviation depends upon how great the data is spread. There are two types of standard deviations in statistics: sample and population standard deviations (Laerd Statistics, n.d.). We calculate the two standard deviations in different ways. The query above is a perfect case of population standard deviation since the data is gotten from an entire populace, the company's employees.
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Although both population and sample standard deviations measure variability, they have qualitative and quantitative differences. In terms of qualitative differences, the sample standard deviation is a statistic determined by using some persons within the population (Taylor, 2019). On the other hand, population standard deviation is a parameter, a specific value determined by each person in the populace (Taylor, 2019). In terms of quantitative differences, whereas the sample standard deviation is divided by n-1, the population standard deviation is divided by n (Taylor, 2019).
The formula for population standard deviation is where, = population standard deviation, = sum of..., = population mean, and n = number of scores.
T he probability that the number of days of vacation taken is less than 10 days
By using the z distribution table and score formula,
z = (number of vacation days-mean vacation days) / vacation days’ standard deviation = -1.33
̵ 1.33 is the probability to the left of z.
The probability is determined by finding the corresponding value of – 1.33 in the z distribution table, that is, P (z < -1.33) = 0.0918 or 9.18%
T he probability that the number of days of vacation taken is more than 21 days
Using
=2.33
The value 2.33 is the probability to the left of z. As such, it is necessary to deduct the probability to the left of z from one to determine probability to the right.
Therefore,
P (z > 2.33) =1 – probability to left of Z
From the z distribution table;
P (z > 2.33) = 1 – 0.9901
P (z > 2.33) = 0.0099 or 0.99%
References
Laerd Statistics, n.d. Standard Deviation. Retrieved 23 October 2020, from https://statistics.laerd.com/statistical-guides/measures-of-spread-standard-deviation.php
Taylor, C. (2019). Differences Between Population and Sample Standard Deviations. Retrieved 23 October 2020, from https://www.thoughtco.com/population-vs-sample-standard-deviations-3126372#:~:text=If%20we%20are%20calculating%20the,the%20number%20of%20data%20values.
