Question

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.

Answer 1

$\mathbf{Solution}$

$\mathbf{given}:\;mean\;(\mu)=14,\;standard \;deviation\;(\sigma )=3$

$\boldsymbol{let\; x \;be\; vacation\; taken\; by \;an \;employee}$

$\mathbf{formula:}\;z=\frac{x-\mu}{\sigma}$

Since sample size 'n' is unknown. You need not to worry whether its sample or population standard deviation. Consider the standard deviation given in the problem as population standard deviation or sample standard deviation. it's not going to affect the answer, Provided if n is unknown.

When to use sample or population standard deviation

The standard deviation is a measure of a score within a set of data. Usually, we are interested in knowing the population standard deviation because our population contains all the values we are interested in. Therefore we normally calculate the population standard deviation if you have the entire population or you have a sample of a larger population. Usually denoted by the letter '$\sigma$'.

You will calculate the sample standard deviation when a small proportion of the sample is drawn from the population. Usually denoted by the letter 's'.

$less \;than \;10\Rightarrow P(x<10)$

$P(x<10)\Rightarrow P\left ( \frac{x-\mu}{\sigma}<\frac{10-14}{3} \right )=P(z<-1.33)$

$\mathbf{from\; z-table, }$

$=\mathbf{{\color{Red} 0.0918}}$

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$more \;than \;21\Rightarrow P(x>21)$

$P(x>21)\Rightarrow P\left ( \frac{x-\mu}{\sigma}>\frac{21-14}{3} \right )=P(z>2.33)$

$\mathbf{from\; z-table, }$

$=\mathbf{{\color{Red} 0.0099}}$

Standard Normal Probabilities Table entry Table entry for z is the area under the standard normal curve to the let of z .01 .03 .04 05 .07 08 .09 3.4 .0003 .0003 .0003 .0ф3 .0003 .0003 .0003 .0003 .0003 .0002 3.2 .0007 .0007 .0006 .006 0005 .0006 .0006 .0005 .0005 .0005 3.0 .0013 .0013 .0013 .012 .0012 .0011 .0011 0 .0010 .0010 -2.8 .0026 .0025 .0024 023 0023 .0022 .0021 0021 .0020 .0019 2.6 0047 0045 0044 043 .0041 0040 0039 0038 0037 0036 2.5 .0062 0060 .0059 057 0055 0054 .0052 0051 .0049 0048 2.3-0099 ggnore negative sign. 7 .0084 -3.3 .0005 .0005 .0005 0004 0004 .0004.0004 .0004 .0003 .0008 .0008 .0008 .0008 .0007 .0007 17 .0016 .0016 .0015 0015 .0014 .0014 -3.1 .0010 .0009 .0009 2.9 .0019 .0018 .0018 -2.7 .0035 .0034 .0033 24 .0082 .0080 .0078.007, 5 0064 2.2 .0139 0136 .0132 2 012 30110 2.1 0179 .0174 0170 .0166 016nC ae TOKIN50143 -2.0 0228 022 .0217 0212 02into area to the right 3 .0183 1.9 .0287 0281 .0274 0268 .026 1.8 0359 0351 0344 0336 0s2of the normal curve 9 .0233 0294 -1.7 .0446 .0436 .0427 0418 040 ..5 .0367 -1.6 .0548 .0537 .0526 .0516 0505 .0495 0485 0475 .0465 .0455 -1.5 .0668 0655 0643 0630 0618 .0606 0594 0582 0571 0559 -14 .0808 0793 .0778 0764 0749 .0735 0721 0708 0694 .0681 -1.3 .0968 .0951 0934 .0918 0901 .0885 .0869 0853 .0838 0823 1.2 1151 1131 1112 1093 .1 1357 1335 13141292 1271 1251 1230 .1210 1190 .1170 1.0 1587 .1562 1539 .1515 1492 1469 .1446 1423 1401 .1379 -0.9 .1841 18141788 1762 1736 .1711 1685 1660 .1635 .1611 -0.8 .2119 2090 .2061 2033 2005 1977 1949 1922 1894.1867 -0.7 2420 2389 2358 2327 2296 2266 .2236 2206 .2177 2148 -0.6 2743 .2709 .2676 .26432611 2578 .2546 25142483 2451 -0.5 3085 3050 3015 2981 2946 .2912 2877 2843 .2810 .2776 -0.4 3446 3409 3372 .3336 3300 3264 3228 3192 .3156 3121 -0.3 3821 37833745 3707 3669 3632 3594 3557 .3520 3483 -0.2 4207 4168 4129 4090 4052 4013 3974 3936 38973859 -0.1 4602 4562 4522 4483 4443 4404 4364 4325 4286 4 4247 -0.0 5000 4960 4920 4880 4840 4801 4761 4721 4681 4641