first gets brailiest michael finds that 35 customers at his grandfather's grocery store use a coupon. to simulate the behavior of the next 5 customers, he writes the numbers 1, 2, 3, 4, and 5 on cards and mixes them up. he writes down that 1, 2, and 3 represent someone using a coupon and 4 and 5 represent someone not using a coupon. michael then randomly selects a card, puts it back, and records the number. he repeats this 5 times to represent 5 customers or 1 trial. he repeats this experiment for a total of 15 trials. the results are shown in the table. 43454 24511 55555 43453 55315 25215 32235 43311 11154 13342 42514 13223 44215 45313 13324 using this simulation, what is the probability that, out of the next 5 customers, 4 or more will use a coupon? enter your answer, as a fraction in simplified form, in the box. first gets brainliest

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Step-by-step explanation:

We are given that,

1, 2, 3 represents someone using a coupon and 4, 5 represents someone not using a coupon.

Now, the results out of 15 trials of the experiment are given to be:

43454, 24511, 55555, 43453, 55315, 25215, 32235, 43311, 11154, 13342, 42514, 13223, 44215, 45313, 13324.

Since, we can see that, the only possibilities where 4 or more customers use coupons are:

32235, 43311, 13342, 13223 and 13324.

That is, out of 15 trials, there are total 5 trials in which 4 or more customers use coupons.

Thus, the probability of 4 or more customers using a coupon is = .

Hence, the required probability is .

sorry not knowing about that

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.