Holly is making

A. What is the probability that this couple spends 45 dollars or more?

guapka [62]

Using the probability table, it is found that:

a) There is a 0.25 = 25% probability that this couple spends 45 dollars or more.

b) The expected amount the couple actually has to pay is $36.85.

Item a:

To find the probabilities involving the total cost, we have to add the variables X and Y from the table, then:

$P(X = 30) = P(X = 15|Y = 15) = 0.2$

$P(X = 35) = P(X = 15|Y = 20) + P(X = 20|Y = 15) = 0.15 + 0.15 = 0.3$

$P(X = 40) = P(X = 15|Y = 25) + P(X = 20|Y = 20) + P(X = 25|Y = 15) = 0.05 + 0.15 + 0.05 = 0.25$

$P(X = 45) = P(X = 20|Y = 25) + P(X = 25|Y = 20) = 0.1 + 0.1 = 0.2$

$P(X = 50) = P(X = 25|Y = 25) = 0.05$

The probability involving values of 45 or more is:

$P(X \geq 45) = P(X = 45) + P(X = 50) = 0.2 + 0.05 = 0.25$

0.25 = 25% probability that this couple spends 45 dollars or more.

Item b:

For a discrete distribution, the expected value is the sum of each outcome multiplied by it's respective probability, hence, involving the 10% discount for prices above $45:

$E(X) = 0.2(30) + 0.3(35) + 0.25(40) + 0.9[0.2(45) + 0.05(50)] = 36.85$

The expected amount the couple actually has to pay is $36.85

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Answer:

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

Step-by-step explanation:

We have to see if 50% = 0.5 is part of the confidence interval.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 1095, \pi = \frac{478}{1095} = 0.4365$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 - 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4118$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4365 + 1.96\sqrt{\frac{0.4365*0.5635}{1095}} = 0.4612$

The 95% confidence interval for the proportion of parents that are satisfied with their children's education is (0.4118, 0.4618). 0.5 is not part of the confidence interval, so this represents evidence that parents' attitudes toward the quality of education have changed.

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