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A network provider investigates the load of its network. The number of concurrent users is recorded at fifty locations (thousands of people), These data are available in data set ConcurrentUsers. (a)Compute the sample mean, variance, and standard deviation of the number of concurrent users. (b)Estimate the standard error of the sample mean. (c)Compute the five-point summary and construct a boxplot. (d)Compute the interquartile range. Are there any outliers

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kendrich

Answer:

see explaination

Step-by-step explanation:

Using the formulla that

sum of terms number of terms sample mean -

Gives the sample mean as \mu=17.954

Now varaince is given by

s^2=\frac{1}{50-1}\sum_{i=1}^{49}(x_i-19.954)^2=9.97

and the standard deviation is s=\sqrt{9.97}=3.16

b) The standard error is given by

\frac{s}{\sqrt{n-1}}=\frac{3.16}{\sqrt{49}}=0.45

c) For the given data we have the least number in the sample is 12.0 and the greatest number in the sample is 24.1

Q_1=15.83, \mathrm{Median}=17.55 and Q_3=19.88

d) Since the interquartile range is Q_3-Q_1=19.88-15.83=4.05

Now the outlier is a number which is greater than 19.88+1.5(4.05)=25.96

or a number which is less than 15.83-1.5(4.05)=9.76

As there is no such number so the given sample has no outliers

a). The sample mean, variance, and standard deviation of the number of concurrent users would be as follows:

Sample mean [tex]=17.954[/tex]

Variance [tex]=9.97[/tex]

Standard deviation [tex]=3.16[/tex]

b). Standard error of the sample mean [tex]=0.45[/tex]

c). Five-point summary would be:

[tex]Q_1=15.83, \mathrm{Median}=17.55 and Q_3=19.88[/tex]

d). Interquartile range [tex]=4.05[/tex]. No. there would be no outliers.

Find the Standard Mean

a). By employing the formula:

Sample Mean = The Sum of the given terms/Number of terms

∵ Sample mean as [tex]\mu=17.954[/tex]

Next,

Variance can be denoted by:

[tex]s^2=\frac{1}{50-1}\sum_{i=1}^{49}(x_i-19.954)^2\\=9.97[/tex]

Next,

Standard deviation:

[tex]s=\sqrt{9.97}\\=3.16[/tex]

b) The standard error can be denoted by:

[tex]\frac{s}{\sqrt{n-1}}=\frac{3.16}{\sqrt{49}}\\=0.45[/tex]

c). For the given data,

The value of the least sample number [tex]= 12.0[/tex]

The value of the greatest number in sample[tex]= 24.1[/tex]

so,

[tex]Q_1=15.83, \\\mathrm{Median}=17.55 \\Q_3=19.88[/tex]

d) The IQ range would be as follows:

[tex]Q_3-Q_1=19.88-15.83\\=4.05[/tex]

Now,

The outlier must be greater than [tex]19.88+1.5(4.05)=25.96[/tex] or less than [tex]15.83-1.5(4.05)=9.76[/tex]

Since there is no number greater or less than the above one, it doesn't contain any outlier.

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