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grigory [225]
3 years ago
13

A researcher surveyed people as they left a museum, asking whether they visited the gift shop and whether they watched the film

about the museum's history. The table shows the relative frequencies of the responses. One person who was surveyed is randomly selected.
What is the probability that the person visited the gift shop, given that the person did not watch the film?

Write the probability as a percent. Round to the nearest tenth if needed. Show all of your work for credit!

Mathematics
1 answer:
BARSIC [14]3 years ago
5 0
Hey there!


P(y) = 0.46

P(x) = 0.49

P(y') = 0.54

P(x') = 0.51

P(y and x') = 0

x = person that watched the film

y = person that visited the gift shop


 P(y and x') / P(x')

0.1 / 0.51 = 0.196 

Now convert that into a percentage!

0.196 = 19.6%

Your answer is 19.6%


Hope I helped!

Let me know if you need anything else!

~ Zoe

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A laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighings. Scale readings in repeated we
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We are given that a laboratory scale is known to have a standard deviation (sigma) or 0.001 g in repeated weighing. Scale readings in repeated weighing are Normally distributed with mean equal to the true weight of the specimen.

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Firstly, the pivotal quantity for 99% confidence interval for the true mean specimen is given by;

        P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample mean weighing of specimen = \frac{3.412+3.416+3.414}{3} = 3.414 g

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            n = sample of specimen = 3

            \mu = population mean

<em>Here for constructing 99% confidence interval we have used z statistics because we know about population standard deviation (sigma).</em>

So, 99% confidence interval for the population​ mean, \mu is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5% level

                                                            of significance are -2.5758 & 2.5758}

P(-2.5758 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X - \mu} < 2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

P( \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.99

<u>99% confidence interval for</u> \mu = [ \bar X-2.5758 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+2.5758 \times {\frac{\sigma}{\sqrt{n} } } ]

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Therefore, 99% confidence interval for this specimen is [3.4125 , 3.4155].

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