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Vera_Pavlovna [14]
4 years ago
8

The number of minutes that Samantha waits to catch the bus is uniformly distributed between 0 and 15 minutes. What is the probab

ility that Samantha has to wait less than 4.5 minutes to catch the bus?
a. 10%
b. 3%
c. 20%
d. 30%
Mathematics
1 answer:
Maslowich4 years ago
5 0

Answer:

option d. 30%

Step-by-step explanation:

Data provided in the question:

a=0

b=15

Now,

The probability density function of the uniform distribution is given as:

f(x)=\frac{1}{15-0}

f(x)=\frac{1}{15}, 0

Therefore,

The required probability will be

P(X

P(X

P(X

P(X

P(X

or

= 0.3 × 100%

= 30%

Hence,

The answer is option d. 30%

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Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
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Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

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

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

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

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