Chris went on a vacation for a week and asked his brother Paul to feed his old cat Charlie. But Paul is forgetful, and Chris is 70% sure Paul will forget to feed his cat. Without food, Charlie will die with probability 0.5.

Answers

Answer 1

Answer:

COMPLETE QUESTION:

Chris went on a vacation for a week and asked his brother Paul to feed his old cat Charlie. But Paul is forgetful, and Chris is 70% sure Paul will forget to feed his cat. Without food, Charlie will die with probability 0.5. With food, he will die with probability 0.03. Chris came back from vacation and found Charlie alive. What is the probability that Paul forgot to feed Charlie (round off to third decimal place)?

Answer:

The probability that Paul forgot to feed charlie is 0.546

Step-by-step explanation:

Lets denote F the event 'Paul forgot to feed Charlie', and L the even 'Charlie is alive', we have

P(F) = 0.7

P(L|F) = 1-0.5 = 0.5

P(L|F^c) = 1-P(L^c|F^c) = 1-0.03 = 0.97

We want to calculate P(F|L). We will use Bayesformula at the start and the theoremoftotalprobability to calculate P(L).

Given that Charlie is alive, the probability that Paul forgot to feed charlie is 0.546.

Answer 2

Answer:

Answer:

P = 0.546

Step-by-step explanation:

Hi,

This is a question of conditional probability, which means to find probability of a situation given that another event has already occured:

$P(A|B) = (P(B|A) P(A))/(P(B)) = (P(A \cap B))/(P(B))$

In this question, we need to find the probability of Charlie being alive if not fed, with the data given below:

$P(Paul\ forgets)= 0.70\nP(Paul\ feeds) = 0.30\nP( Charlie\ dies\ given\ that\ Paul\ forgets) = 0.50\nP( Charlie\ dies\ given\ that\ Paul\ feeds) = 0.03\n$

From this data, we can infer the following:

The probability of Charlie staying alive in both cases:

$P(Charlie\ stays\ alive) = (0.97 * 0.30) + (0.5 * 0.7)\nP(Charlie\ stays\ alive) = 0.641$

We need to find the probability when not fed:

$P (Charlie\ alive\ when\ not\ fed) = (P( dies | not fed) * P(Paul forgets) )/(P(Charlie\ stays\ alive))$

(Remember this is the variation of the conditional probability formula as per our requirement in this question).

$P(Charlie\ alive\ when\ not\ fed) = ((0.5 * 0.7))/(0.641) = 0.546$

Hence, the probability of Charlie being alive when Paul forgets is 0.546.

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Step-by-step explanation:

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Amir went on a bike ride of 60 miles. He realized that if he had gone 5 mph faster, he would have arrived 6 hours sooner. How fast did he actually ride? Amir rode mph on his trip.

Answers

Step-by-step explanation:

distance / rate = time

x = original speed mph

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Data taken from a random sample of 60 students chosen from the student population of a large urban high school indicated that 36 of them planned to pursue post-secondary education. An independent random sample of 50 students taken at a neighboring large suburban high school resulted in data that indicated that 31 of those students planned to pursue post-secondary education. Do these data provide sufficient evidence at the 5% level to reject the hypothesis that these population proportions are equal

Answers

Answer:

No, these data do not provide sufficient evidence at the 5% level to reject the hypothesis that these population proportions are equal.

Step-by-step explanation:

We are given that data taken from a random sample of 60 students chosen from the student population of a large urban high school indicated that 36 of them planned to pursue post-secondary education.

An independent random sample of 50 students taken at a neighboring large suburban high school resulted in data that indicated that 31 of those students planned to pursue post-secondary education.

Let $p_1$ = population proportion of students of a large urban high school who pursue post-secondary education.

$p_2$ = population proportion of students of a large suburban high school who pursue post-secondary education.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that these population proportions are equal}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that these population proportions are not equal}

The test statistics that would be used here Two-sample z proportionstatistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of students of a large urban high school who pursue post-secondary education = $(36)/(60)$ = 0.60

$\hat p_2$ = sample proportion of students of a large urban high school who pursue post-secondary education = $(31)/(50)$ = 0.62

$n_1$ = sample of students of a large urban high school = 60

$n_2$ = sample of students of a large suburban high school = 50

So, the test statistics = $\frac{(0.60-0.62)-(0)}{\sqrt{(0.60(1-0.60))/(60)+(0.62(1-0.62))/(50) } }$

= -0.214

The value of z test statistics is -0.214.

Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that these population proportions are equal.

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