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FrozenT [24]
3 years ago
5

Susan has been on a bowling team for 14 years. After examining all of her scores over that period of time, she finds that they f

ollow a normal distribution. Her average score is 225, with a standard deviation of 13. If during a typical week Susan bowls 16 games, what is the probability that her average score for the week is between 220 and 228? Multiple Choice 0.0618 0.2390 0.7600 0.8212
Mathematics
1 answer:
elixir [45]3 years ago
3 0

Answer:

Step-by-step explanation:

Let X be the random variable that represents a score of Susan. We know that X has a normal distribution with average score of 225 and a standard deviation of 13. For n = 16 games, the average score \bar{X} has a normal distribution with mean of 225 and a standard deviation of 13/\sqrt{16}=13/4=3.25. Besides, the z-score related to 220 is (220-225)/3.25=-1.5385, and the z-score related to 228 is (228-225)/3.25= 0.9231. Therefore, we are looking the probability P(220 < \bar{X} < 228) = P(-1.5385 < Z < 0.9231) = P(Z < 0.9231) - P(Z < -1.5385) = 0.76

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Cost of medium iced caramel latte = $3.59

Step-by-step explanation:

Assume;

Cost of medium iced caramel latte = x

Cost of blueberry muffin = y

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On Thursday

x + y = 5.18......eq1

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2x + 3y = 11.95.....eq2

Eq1 × 3

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Step-by-step explanation:

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Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
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Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

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Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

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3 years ago
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