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kykrilka [37]
3 years ago
14

Suppose that only 25% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

n no other cars are visible. What is the probability that, of 15 randomly chosen drivers coming to an intersection under these conditions, exactly 5 will come to a complete stop?
Mathematics
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

X \sim Binom(n=15, p=0.25)

For this case we can use the probability mass function and we got:

P(X= 5) = (15C5) (0.25)^{5} (1-0.25)^{15-5}= 0.165

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=15, p=0.25)

For this case we can use the probability mass function and we got:

P(X= 5) = (15C5) (0.25)^{5} (1-0.25)^{15-5}= 0.165

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The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev
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Answer:

Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

Also, a sample of four metal sheets is randomly selected from a batch.

Let X bar = Average length of a sheet

The z score probability distribution for average length is given by;

                Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 30.05 inches

           \sigma   = standard deviation = 0.2 inches

             n = sample of sheets = 4

So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) - P(X bar <= 30.25)

P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

 P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 - 0.97725

                                                                                       = 0.0214

                                       

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3 years ago
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So if m=-3 and b=-5, the equation is y = -3x -5
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width=a
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w=5(2legnth)
this doesn't make sense since the legnth is normally longer than the width, but we'll stick with that
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legnth=8.96
width=89.56
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3 years ago
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