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Ostrovityanka [42]

3 years ago

14

Foreign-language study. Choose a student in a U.S. public high school at random and ask if he or she is studying a language othe

r than English. Here is the distribution of results: Language Spanish French German All others None Probability 0.30 0.08 0.02 0.03 0.57 Explain why this is a legitimate probability model. What is the probability that a randomly chosen student is studying a language other than English? What is the probability that a randomly chosen student is studying French, German, or Spanish?

1 answer:

Bezzdna [24]3 years ago

5 0

Answer:

Total probability is 1

0.30+0.08+0.02+0.03+0.57 = 1

P(other than English) = 1-0.57 = 0.43

P(F, G or S) = 0.3+0.08+0.02 = 0.40

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A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

3 years ago

How can the following fraction 33/77 be in simplest form??

Dafna11 [192]

3/7 top and bottom are divisible by 12

8 0

4 years ago

Read 2 more answers

If a number a is chosen at random from the set

Ivanshal [37]

Answer:

D. 3/19

Step-by-step explanation:

12/1 = 12

12/2 = 6

12/3 = 4

These are the only ones that make the equation true.

There are also only 19 numbers because they need to be less than 20 like the question indicated. Therefore, the answer is D.

5 0

3 years ago

HELlPpPp BRAINLIEST thank you <3<br><br> m – 9 = 11

german

Answer:

m=20

Step-by-step explanation:

m-9=11

m-9+9=11+9

m=20

5 0

3 years ago

Read 2 more answers

The product of nine and a number is added to six the result is 24 what is the number

Andru [333]

Hello!

We can make an equation based on what we know

The product of 9 and a number is

9 * x

Added to six

9 * x + 6

Gives the result of 24

9 * x + 6 = 24

Now you solve it algebraically

Subtract 6 from both sides

9 * x = 18

Divide both sides by 9

x = 2

The answer is 2

Hope this helps!

8 0

3 years ago