Find the probability of getting 2 hearts if the cards are drawn from a regular deck and the first card is not replaced.

A true/false quiz consists of twenty questions. If a student answers the twenty questions by making random guesses, what is the probability of getting at least one correct answer?

In how many ways can seven books be arranged on a shelf?

Assume that the cure rate for a certain drug is 20% and that 15 patients are tested with this drug. Find the following:

a) Mean

b) Standard Deviation

c) P (exactly 4 cures)

d) P (exactly 4 cures)

A snack company distributes chips in bags labeled 6 ounces. The local Bureau of Weights and Measures randomly selects 50 bags of chips and obtains a sample mean of 5.9 ounces. Assuming that the standard deviation is known to be 0.8, test the claim at the .05 significance level that the bags contain less than 6 ounces. Identify the following:

a) The Null Hypothesis

b) Critical Value

c) Test Statistics

d) Your Conclusion regarding the Null Hypothesis

e) Your Statement regarding the claim.

Take a shot. What do you think the answers are?

8 months ago

## To find the probability of getting 2 hearts if the cards are drawn from a regular deck and the first card is not replaced, you need to consider the number of favorable outcomes (drawing 2 hearts) and divide it by the total number of possible outcomes.

The number of favorable outcomes can be calculated by multiplying the probability of drawing a heart (13 hearts out of 52 cards) by the probability of drawing another heart from the remaining cards without replacement (12 hearts out of 51 cards).

So, P(2 hearts) = (13/52) * (12/51) = 3/52.

The probability of getting at least one correct answer in a true/false quiz, when a student answers the twenty questions by making random guesses, can be found using the complement rule.

The probability of not getting any correct answer is (1/2)^20, as there are two choices (true/false) for each question.

Therefore, the probability of getting at least one correct answer is 1 - (1/2)^20.

To find the number of ways seven books can be arranged on a shelf, you can use the concept of permutations.

The number of ways to arrange seven books is equal to 7 factorial (7!).

Therefore, there are 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040 ways to arrange the books.

To find the mean, standard deviation, and probabilities for the given drug test scenario, let's break down each part separately.

a) The mean can be calculated by multiplying the cure rate (20%) by the number of patients tested (15). The mean is given by Mean = 0.20 * 15 = 3.

b) The standard deviation can be calculated using the formula: Standard Deviation = sqrt(p(1-p)*n), where p is the cure rate (0.20) and n is the number of patients tested (15). Plugging in the values, we get Standard Deviation = sqrt(0.20 * 0.80 * 15) = sqrt(2.4) â‰ˆ 1.55.

c) P(exactly 4 cures) can be calculated using the binomial probability formula: P(X=k) = (nCk) * p^k * (1-p)^(n-k), where n is the number of trials (15), p is the cure rate (0.20), and k is the number of cures (4). Plug in the values to calculate P(exactly 4 cures).

d) P(exactly 4 cures) can be calculated similarly as in part c, but with k = 5 instead of 4.

For the snack company's test claim, we can perform a hypothesis test using a significance level (Î±) of 0.05.

a) The null hypothesis (H0) would state that the bags contain 6 ounces.

b) The critical value is obtained from the t-distribution table or calculator at the given significance level and degrees of freedom. Since the sample size is 50, the degrees of freedom would be (50 - 1) = 49.

c) The test statistic can be calculated using the formula: Test Statistic = (Sample Mean - Population Mean) / (Standard Deviation / sqrt(n)), where the sample mean is given as 5.9 ounces, the population mean is 6 ounces, the standard deviation is 0.8, and n is the sample size (50).

d) The conclusion regarding the null hypothesis depends on comparing the test statistic with the critical value. If the test statistic falls in the rejection region (i.e., it is less than the critical value), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

e) The statement regarding the claim would depend on the conclusion obtained in d). If the null hypothesis is rejected, it would support the claim that the bags contain less than 6 ounces; otherwise, it would not provide enough evidence to support the claim.

Without specific numerical values for parts c) and d) of the drug test scenario and parts b), c), d), e) of the snack company's test claim, it is not possible to provide the exact answers.

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