Mathematics, 26.03.2020 22:46 lululoveee3433

Ixteen friends meet in Indianapolis for a fun weekend. The distances traveled are asfollows (in miles): 2, 11, 25, 53, 61, 62, 70, 70, 70, 78, 80, 99, 109, 110, 125, 250.(a) Find the sample mean and sample standard deviation.(b) Find the 5-point summary and sketch a boxplot.(c) Sketch a reasonable histogram and the corresponding relative frequency histogram.(d) Describe the data set based on all of the above.

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Answer from: natalie407888

The answers are given below with the explanation of each one.

Step-by-step explanation:

We have sixteen friends.

The distances traveled(in miles) are:

2, 11, 25, 53, 61, 62, 70, 70, 70, 78, 80, 99, 109, 110, 125, 250

We need the frequency of the each data.

Distance - Frequency

2 - 1

11 - 1

25 - 1

53 - 1

61 - 1

62 - 1

70 - 3

78 - 1

80 - 1

99 - 1

109 - 1

110 - 1

125 - 1

250 - 1

(a) Find the sample mean and sample standard deviation.

The mean is represented with sigma symbol (μ) and its obtain by the following equation:

$\mu = \frac{x1+x2+...+xn}{N}$

$\mu = \frac{2+11+25+53+61+62+70+70+70+78+80+99+109+110+125+250}{16}\\\\ \mu=\frac{1230}{16}\\\\\mu = 76.875$

The standard deviation is:

$\sigma = \sqrt{ \Sigma \frac{(\mu -xi)^{2} *f }{n}}$

$\Sigma \frac{(\mu -xi)^{2} *f }{n}= \frac{1}{16}[1(76.875-2)^{2} +1(76.875-11)^{2}+1(76.875-25)^{2}\\\\+ 1(76.875-53)^{2}+ 1(76.875-61)^{2}+ 1(76.875-62)^{2}+ \\\\3(76.875-70)^{2}+ 1(76.875-78)^{2}+ 1(76.875-80)^{2}+ 1(76.875-99)^{2}\\\\ + 1(76.875-109)^{2}+ 1(76.875-110)^{2}+ 1(76.875-125)^{2}+ 1(76.875-250)^{2}]$

$\Sigma \frac{(\mu -xi)^{2} *f }{n}= \frac{1}{16}[48740] = 3046.25$

$\sigma =\sqrt{3046.25}= 55.19$

So the mean and the standard deviation of the sample are:

$\mu=76.875$

$\sigma =55.19$

(b) Find the 5-point summary and sketch a boxplot.

The minimum data, de medium data, the maximum data, the first quarter data and the third quarter data conformed the 5-point summary:

min=2

max=250

The medium is the data of the center of the whole sample. This sample is 16 size so the center are the data number 8 and 9, we need the mean of those two numbers, in this particular case, those two numbers are 70 so we don't need the mean, we can easily conclude that the medium is 70:

2, 11, 25, 53, 61, 62, 70, 70, 70, 78, 80, 99, 109, 110, 125, 250

med=70

The 1st quarter is the center data of the first half of the sample:

2, 11, 25, 53, 61, 62, 70, 70

In this case, we need the mean of those two numbers:

$1stq=\frac{53+61}{2} = 57$

The 3rd quarter is the center data of the second half of the sample:

70, 78, 80, 99, 109, 110, 125, 250

In this case, we need the mean of those two numbers:

$3rdq=\frac{99+109}{2} = 104$

So the 5-point summary is:

min=2

1stq=57

med=70

3rdq=104

max=250

For the box-plot we do a graph with an axis that includes enough data points that include the min and the max. For this case, we want an axis that goes from 2 to 250 and then we put a vertical line by each of the numbers of the 5-point summary. So we put one at 2, another one at 57, one at 70, another one at 104 and one at 250. we connect the three center line into a box, and then we connect the minimum and the maximum to lines. The image of the box-plot is added below in figure 1:

The histogram is added below in figure 2.

The relative frequency histogram is added below in figure 3. For this historigram we need the relative frequency which is obtain by:

$relative frequency = \frac{frecuency}{sum of all frequencies}$

We have only two frequencies 1 and 3

the sum of all frequencies is equal to = 16

$relative frequency1=\frac{1}{16}=0.0625$

$relative frequency 2=\frac{3}{16}=0.1875$

(d) Describe the data set based on all of the above.

The most frequent distance was 70 miles. The rank of the sample is equal to 248, which is relatively wide. And most of the samples are less than 70 miles away from the meeting point, this is reflected on the mean which is 76.875 miles.

Answer from: Quest

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: