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Lilit [14]
3 years ago
9

Fifty draws are made at random with replacement from the box [ 0 0 1 1 1]. There are 33 ticket 1’s among the draws. The expected

value for the sum is ____ , the observed value is ____ , the chance error is ____ , and the standard error is ____ . Fill in the blanks. Hint: Chance error = ObservedOutcome – ExpectedOutcome.
Mathematics
1 answer:
Aleks04 [339]3 years ago
5 0

Answer:

- The Expected value for the sum is 30.

- The Observed sum of 50 draws is 33.

- The Chance Error on the 50 draws is 3.

- The Standard Error on 50 draws is 2.191.

Step-by-step explanation:

The box contains [0, 0, 1, 1, 1]

Using probability to predict the expected outcome.

On one draw, the probability of drawing a 0 is (2/5).

And the probability of drawing a 1 is (3/5).

Probability mass function would look like

X | P(X)

0 | 0.40

1 | 0.60

So, expected value on one draw would be

E(X) = Σ xᵢpᵢ

xᵢ = each variable

pᵢ = probability of each variable

E(X) = (0×0.40) + (1×0.60) = 0.60.

Standard error on one draw = √[Σ(xᵢ - μ)²/N]

μ = E(X) = 0.60

Σ(xᵢ - μ)² = (0 - 0.60)² + (0 - 0.60)² + (1 - 0.6)² + (1 - 0.6)² + (1 - 0.6)² = 1.20

SE = √(1.2/5) = 0.490

So, for 50 draws (with replacement),

E(50X) = 50E(X) = 50 × 0.60 = 30.

For 50 draws, standard error = √50 × 0.490 = 2.191

The expected value for the sum = 30

The observed valued for the sum = (33×1) + (17×0) = 33

Chance Error = (Observed Outcome) - (Expected Outcome) = 33 - 30 = 3

Standard error gives an idea of how large the chance error would be.

Standard error on 50 draws = 2.191

Hope this Helps!!!

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Step-by-Step Explanation:

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Answer:

The differential equation will be like the one shown below

Step-by-step explanation:

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Read 2 more answers
How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?
defon
\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1

(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1

\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow  \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2

\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3

\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}



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4 years ago
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