A network provider investigates the load of its network. The number of concurrent users is recorded at fifty locations (thousands of people)

Question

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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Cora 3 weeks 2022-01-01T18:32:11+00:00 1 Answer 0 views 0

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    2022-01-01T18:33:23+00:00

    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

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45:7+7-4:2-5:5*4+35:2 =? ( )