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3 years ago

10

When would you need to use a small standard deviation?

1 answer:

docker41 [41]3 years ago

4 0

Answer:

You can also use standard deviation to compare two sets of data. For example, a weather reporter is analyzing the high temperature forecasted for two different cities. A low standard deviation would show a reliable weather forecast.

Step-by-step explanation:

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Help me asap …………..p

borishaifa [10]

Answer:

1.) Tan C = $\frac{12}{5}$

2.) Sin Z = $\frac{24}{26}$ or $\frac{12}{13}$ if reduced

3.) Cos A = $\frac{15}{17}$

4.) Cos A = $\frac{12}{37}$

Step-by-step explanation:

Let's keep in mind that;

Sin = $\frac{opposite}{hypotenuse}$

Cos = $\frac{adjacent}{hypotenuse}$

Tan = $\frac{opposite}{adjacent}$

Hope this helps!

5 0

2 years ago

Trish runs 3 miles every 50 minutes. At this rate, how many minutes did Trish tun 12 miles? plzzz help me!!!

Alika [10]

50 x 4 = 200 minutes

The reason why I multiplied by 4 was because 12 divide 3 = 4

200 minutes is equal to 3 hours 20 minutes
There's your answer

Hope this helps!

7 0

3 years ago

Read 2 more answers

The table shows the time Merrida spent driving and the number of miles she drove. She drove the same number of miles each hour.

Oksi-84 [34.3K]

Answer:

35 miles per hour

Step-by-step explanation:

175 -140 = 35

35/1 = 35

6 0

3 years ago

Can someone please help me

viktelen [127]

The surface area Is 98

8 0

3 years ago

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago