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IgorLugansk [536]
2 years ago
7

The manager of a grocery store selected a random sample of 100 customers to estimate the average checkout time. The 90% confiden

ce interval estimate for the average checkout time at the score is[1, 5] minutes. The correct confidence interval interpretation is ___________.
a. 100.0 to 200.0
b. If random samples of 100 customers are repeatedly selected, then that the average checkout time will be between 1 and 5 minutes for 95% of the samples.
c. We estimate with 95% confidence that the average checkout time for the sample of 100 customers is between 1 and 5 minutes.
d. None of these alternatives is correct.
e. We estimate with 95% probability that the average checkout time at the store is between 1 and 5 minutes.
Mathematics
1 answer:
lina2011 [118]2 years ago
3 0

Answer:

c

Step-by-step explanation:

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gtnhenbr [62]
Scientific notation is a way of writing numbers in a form of a decimal multiplied by an exponent of 10.
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Bob has a standard deck of playing cards. If he randomly draws a J, K, 2, and 2, what is the probability that the next card he d
nlexa [21]

Answer:

The correct answer has already been given (twice). I'd like to present two solutions that expand on (and explain more completely) the reasoning of the ones already given.


One is using the hypergeometric distribution, which is meant exactly for the type of problem you describe (sampling without replacement):


P(X=k)=(Kk)(N−Kn−k)(Nn)


where N is the total number of cards in the deck, K is the total number of ace cards in the deck, k is the number of ace cards you intend to select, and n is the number of cards overall that you intend to select.


P(X=2)=(42)(480)(522)


P(X=2)=61326=1221


In essence, this would give you the number of possible combinations of drawing two of the four ace cards in the deck (6, already enumerated by Ravish) over the number of possible combinations of drawing any two cards out of the 52 in the deck (1326). This is the way Ravish chose to solve the problem.


Another way is using simple probabilities and combinations:


P(X=2)=(4C1∗152)∗(3C1∗151)


P(X=2)=452∗351=1221


The chance of picking an ace for the first time (same as the chance of picking any card for the first time) is 1/52, multiplied by the number of ways you can pick one of the four aces in the deck, 4C1. This probability is multiplied by the probability of picking a card for the second time (1/51) times the number of ways to get one of the three remaining aces (3C1). This is the way Larry chose to solve the this.

Step-by-step explanation:


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Answer:

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