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Paul [167]
3 years ago
7

Find the indicated conditional probability

Mathematics
2 answers:
andrezito [222]3 years ago
5 0

Given:

The two way table.

To find:

The conditional probability of P(Drive to school | Senior).

Solution:

The conditional probability is defined as:

P(A|B)=\dfrac{P(A\cap B)}{P(B)}

Using this formula, we get

P(\text{Drive to school }|\text{ Senior})=\dfrac{P(\text{Drive to school and senior})}{P(\text{Senior})}                      ...(i)

From the given two way table, we get

Drive to school and senior = 25

Senior = 25+5+5

           = 35

Total = 2+25+3+13+20+2+25+5+5

         = 100

Now,

P(\text{Drive to school and senior})=\dfrac{25}{100}

P(\text{Senior})=\dfrac{35}{100}

Substituting these values in (i), we get

P(\text{Drive to school }|\text{ Senior})=\dfrac{\dfrac{25}{100}}{\dfrac{35}{100}}

P(\text{Drive to school }|\text{ Senior})=\dfrac{25}{35}

P(\text{Drive to school }|\text{ Senior})=0.7142857

P(\text{Drive to school }|\text{ Senior})\approx 0.71

Therefore, the required conditional probability is 0.71.

Alenkasestr [34]3 years ago
4 0

Answer:

.71

Step-by-step explanation:

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Find the component form of the vector that translates P(4,5) to p'.
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Step-by-step explanation:

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2 years ago
In a quality control test of parts manufactured at Dabco Corporation, an engineer sampled parts produced on the first, second, a
sammy [17]

Answer:

p value = 0.0174

Conclusion : we reject the null hypothesis

Step-by-step explanation:

Thinking step:

We need to perform a test to determine if the proportion of the good parts is the same for all three shifts at a significance level of \alpha = 0.05

Assumption : all the population for each or the three shifts is not equal.

Calculation:

Let p₁ be the sample of the first shift

     p₂ be the sample of the second shift

     p₃ be the sample of the third shift

According to the null hypothesis

H₀ = p₁ = p₂ = p₃

In other words, all the population sample proportions are equal.

Alternatively, we can assume that the three shift are not equal p₁ ≠ p₂ ≠ p₃

Tabulating and performing the \chi² test gives 8.10

degrees of freedom:

df = k - 1

   = 3 - 1 = 2

Thus the degree of freedom is 2

Solving using the MINITAB software gives: p = 0.174

The solution shows that the p value < level of significance, then p-value lies in the range 0.0174≤\alpha≤0.05

Therefore, we reject the null hypothesis based on the fact that the three shifts are not equal.

5 0
3 years ago
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