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ZanzabumX [31]
3 years ago
9

The Orchard Cafe has found that about 3% of the diners who make reservations don't show up. If 83 reservations have been made, h

ow many diners can be expected to show up? Find the standard deviation of this distribution. (Round your answers to two decimal places.) μ = σ =
Mathematics
1 answer:
daser333 [38]3 years ago
6 0

Answer:

\mu = 80.51\\\sigma = 1.56

Step-by-step explanation:

We are given the following information in the question:

Percentage of of the diners who make reservations don't show up = 3%

Number of reservations = 83

Thus, we are given a binomial distribution with n = 83 and p = 0.97

X\sim \text{ Binom}(p = 0.97, n = 83)

\mu = np = (83)(0.97) = 80.51

Around 81 people  can be expected to show up.

\sigma = \sqrt{np(1-p)} = \sqrt{(83)(0.97)(1-0.97)} = 1.5541 \approx 1.56

The standard deviation of this distribution is 1.56

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The hourly median power (in decibels) of received radio signals transmitted between two cities
trasher [3.6K]

Using the lognormal and the binomial distributions, it is found that:

  • The 90th percentile of this distribution is of 136 dB.
  • There is a 0.9147 = 91.47% probability that received power for one of these radio signals is  less than 150 decibels.
  • There is a 0.0065 = 0.65% probability that for  6 of these signals, the received power is less than 150 decibels.

In a <em>lognormal </em>distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{\ln{X} - \mu}{\sigma}

  • It measures how many standard deviations the measure is from the mean.  
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

  • The mean is of \mu = 3.5.
  • The standard deviation is of \sigma = \sqrt{1.22}

Question 1:

The 90th percentile is X when Z has a p-value of 0.9, hence <u>X when Z = 1.28.</u>

Z = \frac{\ln{X} - \mu}{\sigma}

1.28 = \frac{\ln{X} - 3.5}{\sqrt{1.22}}

\ln{X} - 3.5 = 1.28\sqrt{1.22}

\ln{X} = 1.28\sqrt{1.22} + 3.5

e^{\ln{X}} = e^{1.28\sqrt{1.22} + 3.5}

X = 136

The 90th percentile of this distribution is of 136 dB.

Question 2:

The probability is the <u>p-value of Z when X = 150</u>, hence:

Z = \frac{\ln{X} - \mu}{\sigma}

Z = \frac{\ln{150} - 3.5}{\sqrt{1.22}}

Z = 1.37

Z = 1.37 has a p-value of 0.9147.

There is a 0.9147 = 91.47% probability that received power for one of these radio signals is  less than 150 decibels.

Question 3:

10 signals, hence, the binomial distribution is used.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

For this problem, we have that p = 0.9147, n = 10, and we want to find P(X = 6), then:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{10,6}.(0.9147)^{6}.(0.0853)^{4} = 0.0065

There is a 0.0065 = 0.65% probability that for  6 of these signals, the received power is less than 150 decibels.

You can learn more about the binomial distribution at brainly.com/question/24863377

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The diversity of algal species in a lake is calculated for a coastal area before and after an oil spill.
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I think it’s A.Indpendent Variable
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One cell phone plan charges 20$ per month plus$0.15 per minute used.
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So for the first one... Let's just use x for months and m for minutes.

(20m)+(.15m)

20 times the number of months plus .15 times the number of minutes

Same for the next one

(35x)+ (.10m)

35 times the number of months +.10 times the minutes.
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Read 2 more answers
And these. I’m desperate help
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7 is A and 8 is also A
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Dimitri rode his bike 32 miles yesterday. He rode 12 4/5 miles before lunch and the rest of the distance after lunch. How far di
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