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yanalaym [24]
3 years ago
15

Assume that​ women's heights are normally distributed with a mean given by mu equals 63.2 in​, and a standard deviation given by

sigma equals 2.1 in. complete parts a and
b.
a. if 1 woman is randomly​ selected, find the probability that her height is between 62.4 in and 63.4 in. the probability is approximately . 1863. ​(round to four decimal places as​ needed.)
b. if 7 women are randomly​ selected, find the probability that they have a mean height between 62.4 in and 63.4 in

Mathematics
1 answer:
JulijaS [17]3 years ago
8 0

a. Your answer is correct.

b. Divide the given value of sigma by √n where n = sample size = 7. Then perform the same calculation you did for part a. The probabiilty is 0.4427.

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Answer:

Step-by-step explanation:

We have to have 2 different equations to solve this.  One equation will represent the number of tickets sold while the other represents the money collected when the tickets were sold.

We know that adult tickets + children tickets = 2200 tickets.

That's the "number of tickets" equation.  Let's call adult tickets "a" and children's tickets "c".  So a + c = 2200

Now if each adult costs $4, then the expression that represents that as a cost is 4a.  If there is 1 adult, the cost is $4(1) = $4; if there are 2 adults, the cost is $4(2) = $8; if there are 3 adults, the cost is $4(3) = $12, etc.

The same goes for the children's tickets.  If each child's ticket is $1.50, then the expression that represents the cost of a child's ticket is 1.5c (we don't need the 0 at the end; it doesn't change anything to drop it off).  The total money brought in from the cost of these tickets was $5050, so

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4(2200 - c) + 1.5c = 5050 and

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a = 700

There were 1500 children's tickets sold and 700 adult tickets sold.

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