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Kamila [148]
3 years ago
15

Florida state university has 14 statistics classes scheduled for its summer 2013 term. one class has space available for 30 stud

ents, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.
a. what is the average class size assuming each class is filled to capacity?
Mathematics
1 answer:
bonufazy [111]3 years ago
3 0
Hi <span>Bladen8960
So the only numbers that matter are the classes. So let's write it out 30, 60, 60, 60, 60, 60, 60, 60, 60, 70, 100, 100, 100, 100. OK so average is mean so to find mean we add all the classes then divide by how many their are so, adding them all together we get 880 students can go into each classroom so now we divide that by 14 how many classes their are which gives us about 62.
If I'm wrong I'm sorry but I hope this helps!</span>
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Answer:

(a) The correct answer is P (CBM) = 0.79.

(b) The probability of selecting an American female who is not red-green color-blind is 0.996.

(c) The probability that neither are red-green color-blind is 0.9263.

(d) The probability that at least one of them is red-green color-blind is 0.0737.

Step-by-step explanation:

The variables CBM and CBW are denoted as the events that an American man or an American woman is colorblind, respectively.

It is provided that 79% of men and 0.4% of women are colorblind, i.e.

P (CBM) = 0.79

P (CBW) = 0.004

(a)

The probability of selecting an American male who is red-green color-blind is, 0.79.

Thus, the correct answer is P (CBM) = 0.79.

(b)

The probability of the complement of an event is the probability of that event not happening.

Then,

P(not CBW) = 1 - P(CBW)

                   = 1 - 0.004

                   = 0.996.

Thus, the probability of selecting an American female who is not red-green color-blind is 0.996.

(c)

The probability the woman is not colorblind is 0.996.

The probability that the man is  not color- blind is,

P(not CBM) = 1 - P(CBM)

                   = 1 - 0.004  

                   = 0.93.

The man and woman are selected independently.

Compute the probability that neither are red-green color-blind as follows:

P(\text{Neither is Colorblind}) = P(\text{not CBM}) \times  P(\text{not CBW})\\ = 0.93 \times  0.996 \\= 0.92628\\\approx 0.9263

Thus, the probability that neither are red-green color-blind is 0.9263.

(d)

It is provided that a one man and one woman are selected at random.

The event that “At least one is colorblind” is the complement of part (d) that “Neither is  Colorblind.”

Compute the probability that at least one of them is red-green color-blind as follows:

P (\text{At least one is Colorblind}) = 1 - P (\text{Neither is Colorblind})\\ = 1 - 0.9263 \\= 0.0737

Thus, the probability that at least one of them is red-green color-blind is 0.0737.

6 0
3 years ago
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