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pashok25 [27]
3 years ago
6

You have a box of chocolates that contains 59 pieces of which 37 are solid chocolate, 15 are filled with cashews, and 7 are fill

ed with cherries. All the candies look exactly alike. You select a piece, eat it, select a second piece, eat it, and finally eat one last piece. Find the probability of selecting a solid chocolate piece followed by two cherry-filled chocolates. Round your answer to three decimal places.
Mathematics
1 answer:
professor190 [17]3 years ago
6 0
The probability of selecting a solid chocolate piece followed by two cherry-filled chocolates, would be 27.387 or as rounded would be 27.4
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a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.  

From the central limit theorem we have the formula:

z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }

a) z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

\therefore P(\bar X\:>\:1224)=0.0082

b) z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00

The area to the left of z=3.00 is 0.9987

\therefore P(\bar X\:

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

P(1200\:

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

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