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Find the probability of each event.

28. One day, 9 babies are born at a hospital. Assuming each baby has an equal chance of being a boy or girl, what is the probability that exactly 4 of the 9 babies are girls?

29. A gambler places a bet on a horse race. To win, he must pick the top 3 finishers in order. 13 horses of equal ability are entered in the race. Assuming the horses finish in random order, what is the probability that the gambler will win his bet?

Answers

Answer 1

Answer:

28. 0.2461 or 24.61%.

29. 0.00058275 or 0.058275%.

Step-by-step explanation:

Question 28:

we can use the binomial probability formula to calculate the probability:

[tex]\boxed{\bold{P(X = k) = (nCk) * p^k * (1 - p)^(n - k)}}[/tex]

Where:

P(X = k) is the probability of getting exactly k successesnCk is the number of combinations of n items taken k at a timep is the probability of a single successn is the total number of trials

In this case, we have n = 9 babies, and each baby has a 50% chance of being a girl (p = 0.5). We want to find the probability that exactly 4 of them are girls (k = 4).

Using the formula, we can calculate the probability as follows:

[tex]\bold{P(X = 4) = (9C4) * (0.5)^4 * (1 - 0.5)^(9 - 4)}[/tex]

Calculating the values:

(9C4) = 126 (0.5)^4 = 0.0625(1 - 0.5)^(9 - 4) = 0.5^5 = 0.03125

Now, we can substitute these values into the formula:

P(X = 4) = 126 * 0.0625 * 0.03125 = 0.2461 or 24.61%

Therefore, the probability that exactly 4 out of 9 babies are girls is approximately 0.2461 or 24.61%.

Question 29:

We need to calculate the number of possible outcomes to calculate this probability, where the gambler correctly predicts the top 3 finishers in order and divides it by the total number of possible outcomes.

The total number of possible outcomes is the number of permutations of 13 horses taken 3 at a time.

This can be calculated as:

[tex]13P3 = \frac{13! }{(13 - 3)! }= \frac{13! }{ 10! }=\frac{13*12*11*10! }{ 10! }= 13 * 12 * 11 = 1,716[/tex]

Now,

To calculate the number of favorable outcomes where the gambler predicts the top 3 finishers correctly, we need to consider that there is only one correct order for the horses to finish.

Therefore, there is only one favorable outcome.

The probability of the gambler winning his bet is given by:

[tex]\boxed{\bold{P\:(winning) = \frac{Number\: of \:favorable\: outcomes }{ Total \:number \:of \:outcomes}}}[/tex]

[tex]P(winning) = \frac{1 }{1,716}=0,00058275 \: or\:0.058275%[/tex]

Therefore, the probability that the gambler will win his bet is approximately 0.00058275 or 0.058275%.

Answer 2

Answer:

28)  0.246 = 24.6%

29)  1/286 = 0.350%

Step-by-step explanation:

Question 28

We can model the given scenario as a binomial distribution.

Binomial distribution

[tex]X \sim \text{B}(n,p)[/tex]

where:

X is the random variable that represents the number of successes.n is the fixed number of independent trials.p is the probability of success in each trial.

Given the probability that a baby is born a girl is 0.5, and the number of babies is 9:

[tex]\boxed{X \sim \text{B}(9,0.5)}[/tex]

where the random variable X represents the number of babies who are girls.

To find the probability that at exactly 4 babies are girls, we need to find P(X = 4).

To do this, we can use the binomial distribution formula:

[tex]\boxed{\displaystyle \text{P}(X=x)=\binom{n}{x} \cdot p^x \cdot (1-p)^{n-x}}[/tex]

Substitute the values of n = 9, p = 0.5 and x = 4 into the formula:

[tex]\begin{aligned}\displaystyle \text{P}(X=4)&=\binom{9}{4} \cdot 0.5^4 \cdot (1-0.5)^{9-4}\\\\&=\dfrac{9!}{4!\:(9-4)!} \cdot 0.5^4 \cdot 0.5^{5}\\\\&=126 \cdot 0.0625 \cdot 0.03125\\\\& = 0.24609375\end{aligned}[/tex]

Therefore, the probability that 4 babies from a sample of 9 babies are girls is 0.246 (3 s.f.) or 24.6%.

We can also use the binomial probability density function of a calculator to calculate P(X = 4).

Inputting the values of n = 9, p = 0.5 and x = 4 into the binomial pdf:

[tex]\text{P}(X=4)=0.24609375[/tex]

Therefore, this confirms that the probability that 4 babies from a sample of 9 babies are girls is 0.246 (3 s.f.) or 24.6%.

[tex]\hrulefill[/tex]

Question 29

To calculate the probability that the gambler will win his bet, we need to determine the number of favorable outcomes (winning combinations) and the total number of possible outcomes.

The gambler wins if he picks the top three horses in any order. There are 6 ways for the three winners to be arranged in the top three.

There are a total of 13 horses in the race.

The number of ways to choose the first-place horse is 13. After the first-place horse is chosen, there are 12 remaining horses, so the number of ways to choose the second-place horse is 12. Finally, after the first two horses are chosen, there are 11 remaining horses, so the number of ways to choose the third-place horse is 11.

Therefore, the total number of possible outcomes is:

[tex]13 \times 12 \times 11 = 1716[/tex]

Therefore, the probability that the gambler will win his bet is:

[tex]\begin{aligned} \sf Probability &=\sf \dfrac{Favorable \;outcomes}{Total\;outcomes}\\\\&=\dfrac{6}{1716}\\\\&=\dfrac{1}{286}\\\\ & \approx0.350\%\; \sf (3\;d.p.)\end{aligned}[/tex]


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Answer:

2. 10^38, or 10 to the power of 38.
3. 10^15, or 10 to the power of 15.

Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Sally rents a life jacket for a one-time fee of $5. She then
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represents the total cost, in dollars, to rent the life jacket
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Answer: The answer for this question is "A": 5 + 15h

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One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About 12% of American adults identify themselves as black. Suppose we take an SRS of 1500 American adults and let X be the number of blacks in the sample. Use a Normal distribution to estimate the probability that the sample will contain between 165 and 195 blacks.

Answers

Solution :

Given :

p = 12% = 0.12

n = 15..

Mean is defined as the product of a sample size n and the probability p such that :

[tex]$\mu_X = np = 1500 \times 0.12$[/tex]

             = 180

Standard deviation may be defined as the square of a product of the sample size n, the probability p and the probability 1-p :

[tex]$\sigma_X = \sqrt{np(1-p)}$[/tex]

[tex]$\sigma_X = \sqrt{1500 \times 0.12 \times (1-0.12)}$[/tex]

     ≈  12.5857

The z-score is given by :

[tex]$z=\frac {x- \mu}{\sigma } $[/tex]

[tex]$z=\frac {165-180} {12.5857 }$[/tex]

 ≈  -1.19  

[tex]$z=\frac {x- \mu}{\sigma } $[/tex]

[tex]$z=\frac{195-180} {12.5857 }$[/tex]

 ≈  1.19  

Now determining the corresponding probability using table  :

P (165 ≤ X ≤ 195 ) = P (-1.19 ≤ Z ≤ 1.19 )

                            = 1-2 x P(Z < -1.19)

                            = 1-2 x 0.1170

                            = 1-0.2340

                            = 0.7660

                             = 76.60%

"76.06%" would be the probability that the sample will contain between 165 as well as 195 blacks.

Given values,

p = 12%

           = 0.12

n = 1500

The mean will be

→ [tex]\mu x = np[/tex]

By putting the values,

       [tex]= 1500\times 0.12[/tex]

       [tex]= 180[/tex]

The standard deviation will be:

→ [tex]\sigma x = \sqrt{np(1-p)}[/tex]

       [tex]= \sqrt{1500\times 0.12\times (1-0.12)}[/tex]

       [tex]= 12.5857[/tex]

The z-score will be:

→ [tex]z = \frac{x- \mu}{\sigma}[/tex]

     [tex]= \frac{165-180}{12.5857}[/tex]

     [tex]= -1.19[/tex]

and,

→ [tex]z = \frac{195-180}{12.5857}[/tex]

     [tex]= 1.19[/tex]

hence,

The probability will be:

→ [tex]P(165 \leq X \leq 195) = P(-1.19 \leq Z \leq 1.19)[/tex]

                                [tex]= 1-2\times P(Z < -1.19)[/tex]

                                [tex]= 1-2\times 0.1170[/tex]

                                [tex]= 1-0.2340[/tex]

                                [tex]= 0.7660[/tex]

                                [tex]= 76.60[/tex] (%)

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Answer:

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Step-by-step explanation:

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Answers

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Answer:

1

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Step-by-step explanation:

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Answer:

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Answers

Answer:

Step-by-step explanation:

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Answers

Step-by-step explanation:

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Insert the values as follows:

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In an August 2012 Gallup survey of 1,012 randomly selected U.S. adults (age 18 and over), 536 said that they were dissatisfied with the quality of education students receive in kindergarten through grade 12.

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Answers

Answer:

Yes, result is significant ; PVALUE < α

Step-by-step explanation:

Given :

x = 536

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H1 : P0 > 0.5

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(Phat - P0) ÷ sqrt[(P0(1 - P0)) / n]

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0.03 ÷ 0.0157173

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Using the Pvalue from test statistic :

Pvalue = 0.02815

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0.02815 < 0.05

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Answers

Answer:

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Step-by-step explanation:

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paddling pool
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Answer:

35

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