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2 years ago

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You want to make five-letter codes that use the letters A, F, E, R, and M without repeating any letter. What is the probability

that a randomly chosen code starts with M and ends with E?

1 answer:

Feliz [49]2 years ago

4 0

Answer:

The probability that a randomly chosen code starts with M and ends with E is 0.05 ....

Step-by-step explanation:

According to the given statement we have to make five letter code from A, F, E, R, and M without repeating any letter. We have to find that what is probability that a randomly chosen code starts with M and ends with E.

Thus the probability of picking the first letter M = 1/5

After that we require the sequence (not E, not E, not E) which is equal to:

= 3/4 * 2/3* *1/2

= 1/4

Now multiply 1/5 and 1/4

1/5 * 1/4

= 1/20

= 0.05

Therefore the probability that a randomly chosen code starts with M and ends with E is 0.05 ....

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From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that ag

Jlenok [28]

Answer:

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

Step-by-step explanation:

1) Data given and notation

$\bar X=18.1$ represent the mean age when smokers first start to smoke varies

$s=1.3$ represent the standard deviation for the sample

$\sigma=2.1$ represent the population standard deviation

$n=37$ sample size

$\mu_o =5.7$ represent the value that we want to test

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

t 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 determine if the mean starting age is at least 19, the system of hypothesis would be:

Null hypothesis: $\mu\geq 19$

Alternative hypothesis: $\mu < 19$

We know the population deviation, so for this case 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:

$z=\frac{18.1-19}{\frac{2.1}{\sqrt{37}}}=-2.607$

Calculate the P-value

Since is a one-side lower test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we to reject the null hypothesis, and the the actual mean starting age of smokers is significantly lower than 19.

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