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3 years ago

Problem 1: Given the following probabilities P(Ac) = 0.45, P(B) = 0.34, and P(B|A) = 0.23. Find the following probabilities:

Mathematics

1 answer:

-BARSIC- [3]3 years ago

8 0

Answer:

P(A)=0.55

P(A and B)=P(A∩B)=0.1265

P(A or B)=P(A∪B)=0.7635

P(A|B)=0.3721

Step-by-step explanation:

P(A')=0.45

P(A)=1-0.45=0.55

P(B∩A)=?

P(B|A)=0.23

P(B|A)=(P(A∩B))/P(A)

0.23=(P(A∩B))/0.55

P(A∩B)=0.23×0.55=0.1265

P(A∪B)=P(A)+P(B)-P(A∩B)

=0.55+0.34-0.1265

=0.7635

P(A|B)=[P(A∩B)]/P(B)=0.1265/0.34 ≈0.3721

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Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, $\overline x _1$ = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, $\overline x _2$ = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

$\left (\bar{x}_{2}- \bar{x}_{1} \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}$

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

$t_{(0.025, \, 18)}$ = 1.734

$C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}$

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

$t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}$

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The two sample t-statistic ≈ 3.43

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Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

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3 years ago

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3 years ago

App 19.study

lions [1.4K]

Answer:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

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Step-by-step explanation:

Given

$n = 125$

See attachment for proper table

Required

Complete the table

Experimental probability is calculated as:

$Pr = \frac{Frequency}{n}$

We use the above formula when the frequency is known.

For result of roll 2, 4 and 6

The frequencies are 13, 29 and 6, respectively

So, we have:

$Pr(2) = \frac{13}{125} = 0.104$

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When the frequency is to be calculated, we use:

$Pr = \frac{Frequency}{n}$

$Frequency = n * Pr$

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The probabilities are 0.144 and 0.296, respectively

So, we have:

$Frequency(3) = 125 * 0.144 = 18$

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For roll of 1 where the frequency and the probability are not known, we use:

$Total \ Frequency = 125$

So:

Frequency(1) added to others must equal 125

This gives:

$Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125$

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$1 \to 22 \to 0.176$

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$3 \to 18 \to 0.144$

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$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

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3 years ago

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V125BC [204]

Answer:

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Step-by-step explanation:

the $\sqrt{-27}$ needs to be simplified

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$\sqrt{-1}$ simplifies to i ;and can be broken down to $\sqrt{9}$ and $\sqrt{3}$

$\sqrt{9}$ can be simplified further to 3

Now all you have to do is put it together

4(which you didnt have to do anything to)+ 3i( 3 from the sqrt of 9 and i from the sqrt of -1) sqrt 3( which cant be simplified further)

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3 years ago

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A similar problem is given at brainly.com/question/6120515

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3 years ago