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2 years ago

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The value of 4 in 34,258 is how many times the value of 4 in 47,614

1 answer:

Zina [86]2 years ago

8 0

The value of 4 in 34 258 is 4000 - four thousand.

The value of 4 on 47614 is 40 000 - forty thousand.

The first is10 times less than the second and the second is ten times more than the first:
4000*10=40000
You can also say that the first is 0.1 times the second.

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A true-false quiz with 10 questions was given to a statistics class. Following is the probability distribution for the score of

disa [49]

Answer:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48$

For this case this value means that the expected score is about 7.48

Step-by-step explanation:

For this case we assume the following probability distribution:

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P(X) 0.05 0.15 0.33 0.28 0.12 0.07

First we need to find the expected value (first moment) and the second moment in order to find the variance and then the standard deviation.

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=5*0.05 +6*0.15 +7*0.33 +8*0.28+ 9*0.12 +10*0.07=7.48$

For this case this value means that the expected score is about 7.48

In order to find the standard deviation we need to find first the second moment, given by :

$E(X^2)=\sum_{i=1}^n X^2_i P(X_i)$

And using the formula we got:

$E(X^2)=(5^2 *0.05)+(6^2 *0.15)+(7^2 *0.33)+(8^2 *0.28)+ (9^2 *0.12 +(10^2 *0.07))=57.46$

Then we can find the variance with the following formula:

$Var(X)=E(X^2)-[E(X)]^2 =57.46-(7.48)^2 =1.5096$

And then the standard deviation would be given by:

$Sd(X)=\sqrt{Var(X)}=\sqrt{1.5096}=1.229$

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Answer:

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From here you have your equations, turn them into the f(x) form.

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