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Phoenix [80]
3 years ago
13

A museum employee surveys a random sample of 350 visitors to the museum. If those visitors, 266 stopped at the gift shop. Based

on these results, about how many people out of 3200 visitors to the museum would be expected to stop at the gift shop?
Mathematics
2 answers:
Free_Kalibri [48]3 years ago
8 0

Answer:

about 2432 people(well exactly this many are supposed to go)

Step-by-step explanation:

266 visitors out of 350 means 19/25 people go to the gift shop. now we multiply this by 3200 to get the number of people who would go to get 2432

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irina [24]3 years ago
7 0

Answer: 2432 visitors

Step-by-step explanation:

An employee of a museum surveyed a random sample of 350 visitors that came to the museum. Of the visitors that were surveyed, 266 stopped at the gift shop.

Since 266 out of 350 stopped at the gist shop, we find the fraction or decimal of people who stopped at the gift shop. This will give 266/350 = 19/25 or 0.76.

If 3200 visitors come to the museum, the expected number of people to stop at the gift shop are:

= 0.76 × 3200

= 2432

2432 visitors are expected to stop at the gift shop out of 3200 visitors.

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15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
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Answer:

1) Probability that all five are good = 0.46

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The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

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26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

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Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

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P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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