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Kazeer [188]
3 years ago
8

Shania's test scores in 8 subjects were 88, 91, 85, 74, 69, 72, 80, and 87. Which measure should Shania use to find how much the

scores on her tests vary, on average, from the mean score?
mode

mean absolute deviation

interquartile range

mean
Mathematics
1 answer:
valina [46]3 years ago
8 0

Answer with explanation:

Test Scores of Shania's in 8 subject are: 88, 91, 85, 74, 69, 72, 80, and 87.

To calculate mean we add all the variate and divide by total number of Variate in the data set.

So, Mean of the Data set

            =\frac{88+ 91+ 85+ 74+69+ 72+ 80+ 87}{8}\\\\=\frac{646}{8}\\\\=80.75

To calculate , how much the scores on her tests vary, on average, from the mean score

We will Calculate mean absolute Deviation.

To calculate it,we will find absolute difference of each variate from mean.

| 88 - 80.75|=7.25

|91-80.75|=10.25

|85 -80.75|=4.25

|74-80.75|=6.75

......

........

Then Shania should calculate mean of the new data set obtained,which is absolute difference of each variate from mean.

→The Measure that,Shania should use to find how much the scores on her tests vary, on average, from the mean score is:

Mean absolute deviation

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3 years ago
Find the cdf F(x) associated with each of the following probability density functions. Sketch the graphs of f(x) and F(x).
wolverine [178]

Answer:

See steps below

Step-by-step explanation:

a)

\bf f(x)=3(1-x)^2\;(0

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The cdf associated with f is

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<h3>See picture 1 </h3>

The median is a point x such that

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so, the median is

\bf 1-(1-x)^3=1/2\rightarrow (1-x)^3=1/2\rightarrow \boxed{x=1-\sqrt[3]{2}}

The 25th percentile equals the 1st quartile and is a point x such

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and the 25th percentile is

\bf 1-(1-x)^3=1/4\rightarrow (1-x)^3=3/4\rightarrow \boxed{x=1-\sqrt[3]{3/4}}

b)

\bf f(x)=\frac{1}{x^2}\;(1

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The cdf associated with f is

\bf \boxed{F(x)=1-\frac{1}{x}} for x>1

<h3>See picture 2 </h3>

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\bf 1-\frac{1}{x}=\frac{1}{2}\rightarrow \boxed{x=2}

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c)

f(x) = 1/3 for 0<x<1 or 2<x<4

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\bf F(x)=\frac{1}{3}+\int_{2}^{x}\frac{dt}{3}=\frac{1}{3}+\frac{x-2}{3}=\frac{x-1}{3}\;(2

The cdf associated with f is

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<h3>See picture 3 </h3>

The median is

\bf \frac{x-1}{3}=1/2\rightarrow x=1+3/2\rightarrow \boxed{x=5/2}

The 25th percentile is  

\bf \frac{x}{3}=1/4\rightarrow \boxed{x=3/4}

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Answer:

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Step-by-step explanation:

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