Question

1 - 3 mean, median, mode, range, standard deviation, and variance part 2 of 2
following are the prices of 12 tickets listed on the ticket racket ticket - broker site for a bruce springsteen concert. (in dollars)
75, 120, 120, 145, 150, 150, 150, 175, 175, 200, 225, 275
round your answers to the nearest hundredth.
what is the mean ticket price?
what is the median ticket price?
what is the range?
what is the variance?
what is the standard deviation?

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 12$ and $x_{i}$ are the data - points. $\sum_{i=1}^{12}x_{i}=75 + 120+120 + 145+150+150+150+175+175+200+225+275=1860$. So, $\bar{x}=\frac{1860}{12}=155$.

Step2: Calculate the median

First, order the data: $75,120,120,145,150,150,150,175,175,200,225,275$. Since $n = 12$ (an even number), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values. The 6th and 7th values are 150 and 150, so the median $=\frac{150 + 150}{2}=150$.

Step3: Calculate the range

The range is the difference between the maximum and minimum values. Range $=275−75 = 200$.

Step4: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
$(75−155)^{2}=(-80)^{2}=6400$, $(120 - 155)^{2}=(-35)^{2}=1225$ (twice), $(145−155)^{2}=(-10)^{2}=100$, $(150−155)^{2}=(-5)^{2}=25$ (thrice), $(175−155)^{2}=20^{2}=400$ (twice), $(200−155)^{2}=45^{2}=2025$, $(225−155)^{2}=70^{2}=4900$, $(275−155)^{2}=120^{2}=14400$.
$\sum_{i = 1}^{12}(x_{i}-\bar{x})^{2}=6400+2\times1225 + 100+3\times25+2\times400+2025+4900+14400=32800$.
$s^{2}=\frac{32800}{11}\approx2981.82$.

Step5: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$. Since $s^{2}\approx2981.82$, $s=\sqrt{2981.82}\approx54.61$.

Answer:

Mean: $155.00$
Median: $150.00$
Range: $200.00$
Variance: $2981.82$
Standard Deviation: $54.61$