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3 years ago

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Two dice are rolled. What is the probability that the sum of the numbers is exactly 10? What is the probability that the sum of

the numbers is at least 10? What is the probability that the sum of the numbers is exactly 10, given that it is at least 10?

1 answer:

never [62]3 years ago

5 0

Answer:

exactly 10: 8.33 %

at least 10: 16.66 %

Step-by-step explanation:

first we have to see the amount of favorable results

4+6 = 10

6+4 =10

5+5=10

then we have to see the amount of possible results

Each dice has 6 faces and there are 2 dice, so if we make 6 ^ 2 we will get the number of possible combinations

6^2 =

6*6 = 36

Now we only do the number of favorable results on the possible results and multiply it by 100 and obtain the probability

100*3/36 = 8.33 %

we do the same for those greater than or equal to 10

4+6 = 10

6+4 = 10

5+5 = 10

6+5 = 11

5+6 = 11

6+6 = 12

100*6/36 = 16.66 %

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Answer:

a

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b

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Step-by-step explanation:

From the question we are told that

The sample size is n = 10

The population mean is $\mu = 13.4 000 \ angstroms$

The level of significance is $\alpha = 0.05$

The sample data is

13.3987, 13.3957, 13.3902, 13.4015, 13.4001, 13.3918, 13.3965, 13.3925, 13.3946, and 13.4002

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Generally the sample standard deviation is mathematically represented as

$\sigma = \sqrt{\frac{\sum (x_i - \= x)^2}{n} }$

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The alternative hypothesis is $H_a : \mu \ne 13.4000$

Generally the test statistics is mathematically represented as

$z = \frac{\= x - \mu}{\frac{\sigma}{\sqrt{n} } }$

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From the z-table $P(z > 3.08)= 0.001035$

So

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So from the obtained value we see that

$p-value < \alpha$

Hence the null hypothesis is rejected

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$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

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=> $E = 0.00318$

Generally the 99% confidence interval is mathematically represented as

$13.3962 - 0.00318 < \mu < 13.3962 + 0.00318$

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