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Juliette [100K]
2 years ago
6

*please help me out if you know how to solve, last person I asked just put a random awnser for point, the problem is bellow and

there is a chart in the picture.*
6. In both Mr. Jacquez's Math 2 classes,
Period 2 and Period 3 were given the same test. The table below shows the results from the two classes.

a. Find the probability that a randomly selected student from Period 2's class failed the test.

b. Find the probability that a randomly selected student who failed the test is from Period 3's class​

Mathematics
1 answer:
kicyunya [14]2 years ago
8 0

     For both of these, we will divide the wanted outcome by the number of possible outcomes. We will end up with a decimal. A decimal multiplied by 100 becomes a percent.

[a] 50%

\displaystyle \frac{\text{wanted outcomes}}{\text{possible outcomes}} =\frac{\text{student failed}}{\text{Period 2's class}} =\frac{12}{24} =0.5,\:0.5*100=50\%

[b] 52%

\displaystyle \frac{\text{wanted outcomes}}{\text{possible outcomes}} =\frac{\text{student failed from Period 3}}{\text{fails from both classes}} =\frac{13}{25} =0.52,

    \displaystyle 0.52*100=52\%

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H 0 : P = 0.10

H 1 : P > 0.10 (Right tailed test)

n = 527, P = 0.1235

The test statistics

x = P - p / \sqrt{p*(1-p) / n}

x = 0.1235 - 0.10 / \sqrt{0.10 * 0.90 / 527}

x = 0.0235 / \sqrt{0.09 / 527}

x = 0.0235 / \sqrt{0.00017077798}

x = 0.0235 / 0.0130682049264618

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