An insurance company crashed four cars in succession at 5 miles per hour. The cost of repair for each of the four crashes was$423, $468, $403, $222 . Compute the range, samplevariance, and sample standard deviation cost of repair.

asigurato7
2022-07-31
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Bradley Sherman

Answered 2022-08-01
Author has **17** answers

Step 1

The range of a data set is the highest data value minus the lowest data value. Here, the range of the cost of repair is $468-222=246$.

The mean of the given data set for the cost of repair is $\frac{423+468+403+222}{4}=\$1516/4=379$.

If there are n data points, then the Variance = sum of the squares of (Data points - mean) divided by $(n-1)$.

Here, the difference of the data points from the mean are $(423-379)=44,(468-379)=89,(403-379)=24\text{}\text{and}\text{}(222-379)=-157$. Hence the variance is $[442+892+242+(-157)2]/3=(1936+7921+576+24649)/3=35082/3=11694$.

Step 2

The standard deviation is the square root of the average squared deviation from the mean.

Here, the standard deviation is $\sqrt{11694}=108.14$ ( on rounding off to 2 decimal places).

The range of a data set is the highest data value minus the lowest data value. Here, the range of the cost of repair is $468-222=246$.

The mean of the given data set for the cost of repair is $\frac{423+468+403+222}{4}=\$1516/4=379$.

If there are n data points, then the Variance = sum of the squares of (Data points - mean) divided by $(n-1)$.

Here, the difference of the data points from the mean are $(423-379)=44,(468-379)=89,(403-379)=24\text{}\text{and}\text{}(222-379)=-157$. Hence the variance is $[442+892+242+(-157)2]/3=(1936+7921+576+24649)/3=35082/3=11694$.

Step 2

The standard deviation is the square root of the average squared deviation from the mean.

Here, the standard deviation is $\sqrt{11694}=108.14$ ( on rounding off to 2 decimal places).

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So my task is the following: Consider a sequence of random variables $({X}_{n}{)}_{n\in \mathbb{N}}$ with ${X}_{n}\stackrel{a.s.}{\to}X$ and ${X}_{n}\le B$ a.s. for $B\in \mathbb{R}$. Show that ${X}_{n}\stackrel{{\mathcal{L}}^{p}}{\to}X$ for all $p\in \mathbb{N}$.

My ideas till now:

1. In a first step i tried to show that the expected value is bounded by both sides from zero: $0\le \mathbb{E}(|{X}_{n}-X{|}^{p})\le \dots 0$ Problem here is that i would need that $|{X}_{n}|\le B$ a.s.

2. My second attempt was to use the dominated convergence theorem. With ${Y}_{n}=B$ it follows that if $Y=B$ that ${Y}_{n}\stackrel{a.s.}{\to}Y$ and $\mathbb{E}(|{Y}_{n}-Y|)$. Now I also need that $|{X}_{n}|\le Y$ and then it follows that $\mathbb{E}(|{X}_{n}-X|)\underset{n\to \mathrm{\infty}}{\to}0$.

Probably someone can give me a hint or another way of solving this.

My ideas till now:

1. In a first step i tried to show that the expected value is bounded by both sides from zero: $0\le \mathbb{E}(|{X}_{n}-X{|}^{p})\le \dots 0$ Problem here is that i would need that $|{X}_{n}|\le B$ a.s.

2. My second attempt was to use the dominated convergence theorem. With ${Y}_{n}=B$ it follows that if $Y=B$ that ${Y}_{n}\stackrel{a.s.}{\to}Y$ and $\mathbb{E}(|{Y}_{n}-Y|)$. Now I also need that $|{X}_{n}|\le Y$ and then it follows that $\mathbb{E}(|{X}_{n}-X|)\underset{n\to \mathrm{\infty}}{\to}0$.

Probably someone can give me a hint or another way of solving this.

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Multiplying whole number with fractions.

I'm looking at a solution to a math problem and there are obviously some rules regarding multiplication of fractions that I don't know.

Can someone make any sense of this?

${s}_{n}=625\cdot \frac{{\left(\frac{1}{5}\right)}^{n}-1}{\frac{1}{5}-1}=625\cdot \frac{{\left(\frac{1}{5}\right)}^{n}-1}{-\frac{4}{5}}=-\frac{3125}{4}\cdot ({\left(\frac{1}{5}\right)}^{n}-1)=\frac{3125}{4}\cdot (1-{\left(\frac{1}{5}\right)}^{n})$

Can someone explain to me how you get from each step to another?

I'm looking at a solution to a math problem and there are obviously some rules regarding multiplication of fractions that I don't know.

Can someone make any sense of this?

${s}_{n}=625\cdot \frac{{\left(\frac{1}{5}\right)}^{n}-1}{\frac{1}{5}-1}=625\cdot \frac{{\left(\frac{1}{5}\right)}^{n}-1}{-\frac{4}{5}}=-\frac{3125}{4}\cdot ({\left(\frac{1}{5}\right)}^{n}-1)=\frac{3125}{4}\cdot (1-{\left(\frac{1}{5}\right)}^{n})$

Can someone explain to me how you get from each step to another?

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