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nata0808 [166]
2 years ago
5

A number which is now a rattional number is a​

Mathematics
1 answer:
Flura [38]2 years ago
6 0

Answer:

??? Incomplete sentence

Step-by-step explanation:

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A survey asked only girls to identify their favorite item on the
MA_775_DIABLO [31]

Answer:

B

Step-by-step explanation:

A biased sample is a sample that does not accurately represent the population.

The survey asked only girls, but we can assume this school has both boys and girls. Therefore, by only asking girls, it creates a bias, because the boys in the school are not represented.

Therefore, the correct choice is B: biased sample.

6 0
2 years ago
When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let
sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = <em>the number of defective boards in a random sample of size, n = 25</em>

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r}  ; x = 0,1,2,......

where, n = number of trials (samples) taken = 25

            r = number of success

            p = probability of success which in our question is percentage

                   of defectivs, i.e. 5%

(a) P(X = 3) =  \binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

                   =  2300 \times 0.05^{3}\times 0.95^{22}

                   =  <u>0.093</u>

(b) P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= \binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

=  1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}

=  <u>0.966</u>

(c) P(X \geq 4) = 1 - P(X < 4) = 1 - P(X \leq 3)

                    =  1 - 0.966

                    =  <u>0.034</u>

<u></u>

(d) P(1 ≤ X ≤ 3) =  P(X = 1) + P(X = 2) + P(X = 3)

=  \binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}

=  25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}

=  <u>0.688</u>

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

     P(X = 0) =  \binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}

                   =  1 \times 1\times 0.95^{25}

                   =  <u>0.277</u>

(f) The expected value of X is given by;

       E(X)  =  n \times p

                =  25 \times 0.05  = 1.25

The standard deviation of X is given by;

        S.D.(X)  =  \sqrt{n \times p \times (1-p)}

                     =  \sqrt{25 \times 0.05 \times (1-0.05)}

                     =  <u>1.089</u>

8 0
2 years ago
Bill furrier marks up minks coats $3000. This represents 50% mark up cost. What is the cost of the coats?
Romashka [77]
1500. since hes marking it up by 50% then take half of 3000 to get your answer
4 0
3 years ago
After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni
antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) p_v =P(z>2.8)=1-P(z

c) Using the significance level assumed \alpha=0.01 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.    

Step-by-step explanation:

1) Data given and notation  

n=100 represent the random sample taken    

X=64 represent were in favor of firing the coach

\hat p=\frac{64}{100}=0.64 estimated proportion for were in favor of firing the coach

p_o=0.5 is the value that we want to test since the problem says majority    

\alpha represent the significance level (no given, but is assumed)    

z would represent the statistic (variable of interest)    

p_v represent the p value (variable of interest)    

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 90% confidence interval the value of \alpha=1-0.9=0.1 and \alpha/2=0.05, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.64

And replacing into the confidence interval formula we got:

0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561

0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :    

Null Hypothesis: p \leq 0.5  

Alternative Hypothesis: p >0.5  

We assume that the proportion follows a normal distribution.    

This is a one tail upper test for the proportion of  union membership.  

The One-Sample Proportion Test is "used to assess whether a population proportion \hat p is significantly (different,higher or less) from a hypothesized value p_o".  

Check for the assumptions that he sample must satisfy in order to apply the test  

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.  

b) The sample needs to be large enough  

np_o =100*0.64=64>10  

n(1-p_o)=100*(1-0.64)=36>10  

Calculate the statistic    

The statistic is calculated with the following formula:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}  

On this case the value of p_o=0.5 is the value that we are testing and n = 100.  

z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8

The p value for the test would be:  

p_v =P(z>2.8)=1-P(z

Part c

Using the significance level assumed \alpha=0.01 we see that p_v so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.    

6 0
3 years ago
Help!! is it $486?
xxTIMURxx [149]
The answer on APEX would be $2430
8 0
3 years ago
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