Question

A local amusement park has 30 rides that park visitors can go on. The following table shows the relative frequency distribution for the number of rides that a park visitor will typically go on during one day at the park. The table also shows the deviation, or difference, from 21, the mean of the distribution. Number of rides and attractions 10 15 20 25 30 Deviation −11 −6 −1 4 9 Relative frequency 0.1 0.2 0.2 0.4 0.1 Which of the following is closest to the standard deviation of the distribution? 5.83 5.83 A 7.07 7.07 B 7.90 7.90 C 20 20 D 34

Answer 1

Answer:

5.83

Step-by-step explanation:

Find the mean:

10(.1) + 15(.2) + 20(.2) + 25(.4) + 3-).1)

u = 21

Find the standard deviation:

$\sqrt{(10 -21)x^{2} (.1) + (15 -21)x^{2} (.2)+(20-21)x^{2} (.2)+(25-21)x^{2} (.4)+(30-21)x^{2} (.1)}$= 5.83

Answer 2

The standard deviation is of 5.83.

The expected value is given by each value multiplied by it's relative frequency, thus:

$E(X) = 0.1(10) + 0.2(15) + 0.2(20) + 0.4(25) + 0.1(30) = 21$

The standard deviation is given by the square root of the sum of the difference squared of each value and the mean, multiplied by it's relative frequency.

Then:

$\sqrt{V(X)} = \sqrt{0.1(10 - 21)^2 + 0.2(15 - 21)^2 + 0.2(20 - 21)^2 + 0.4(25 - 21)^2 + 0.1(30 - 21)^2} = 5.83$

The standard deviation is of 5.83.

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