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4 years ago

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find the standard deviation of binomial random variable. A die is rolled 18 times and the number of fours that comes up is talli

ed

1 answer:

raketka [301]4 years ago

4 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The standard deviation is $\sigma =1.5811$

Step-by-step explanation:

The sample size is n = 18

Generally the probability of getting a four in the toss of the fair die is mathematically represented as

$p = \frac{1}{6 }$

While the probability of not getting a four is

$q = 1 - p$

$q = 1 - \frac{1}{6}$

$q = \frac{5}{6}$

Now the standard deviation for the binomial random number is mathematically represented as

$\sigma = \sqrt{n * pq }$

substituting values

$\sigma = \sqrt{18 * \frac{1}{6}* \frac{5}{6} }$

$\sigma =1.5811$

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3 years ago

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3 years ago

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Answer:

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Step-by-step explanation:

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$\frac{8}{b} =\frac{2\frac{2}{3} }{3}\\\\\mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 2\frac{2}{3}=\frac{8}{3}\\\\\frac{8}{b}=\frac{\frac{8}{3}}{3}\\\\8\cdot \:3=b\frac{8}{3}\\\\24=b\frac{8}{3}\\\\8b=72\\\\b=9$

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3 years ago

A researcher would like to determine whether a new tax on legalized marijuana has had any effect on people’s purchasing behavior

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Answer:

a) $z=\frac{386-410}{\frac{60}{\sqrt{9}}}=-1.2$

$p_v =2*P(z$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not different from 410 at 5% of signficance.

b) $z=\frac{386-410}{\frac{30}{\sqrt{9}}}=-2.4$

Since is a two-sided test the p value would be:

$p_v =2*P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly different from 410 at 5% of signficance.

Step-by-step explanation:

Part a

Data given and notation

$\bar X=386$ represent the sample mean

$\sigma=60$ represent the population standard deviation

$n=9$ sample size

$\mu_o =410$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean differs from 410, the system of hypothesis would be:

Null hypothesis: $\mu =410$

Alternative hypothesis: $\mu \neq 410$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{386-410}{\frac{60}{\sqrt{9}}}=-1.2$

P-value

Since is a two-sided test the p value would be:

$p_v =2*P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that the true mean is not different from 410 at 5% of signficance.

Part b

$z=\frac{386-410}{\frac{30}{\sqrt{9}}}=-2.4$

P-value

Since is a two-sided test the p value would be:

$p_v =2*P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significantly different from 410 at 5% of signficance.

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3 years ago