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tino4ka555 [31]
3 years ago
15

Assume that the ages for first marriages are normally distributed with a mean of 26 years and a standard deviation of 4 years. W

hat is the probability that a person getting married for the first time is in his or her twenties? (Round your answer to 4 decimal places.)
Mathematics
1 answer:
Snowcat [4.5K]3 years ago
6 0

Answer:

0.7743

Step-by-step explanation:

Mean of age = u = 26 years

Standard Deviation = \sigma = 4 years

We need to find the probability that the person getting married is in his or her twenties. This means the age of the person should be between 20 and 30. So, we are to find P( 20 < x < 30), where represents the distribution of age.

Since the data is normally distributed we can use the z distribution to solve this problem. The formula to calculate the z score is:

z=\frac{x-u}{\sigma}

20 converted to z score will be:

z=\frac{20-26}{4}=-1.5

30 converted to z score will be:

z=\frac{30-26}{4}=1

So, now we have to find the probability that the z value lies between -1.5 and 1.

P( 20 < x < 30) = P( -1.5 < z < 1)

P( -1.5 < z < 1 ) = P(z < 1) - P(z<-1.5)

From the z-table:

P(z < 1) = 0.8413

P(z < -1.5) =0.067

So,

P( -1.5 < z < 1 ) = 0.8413 - 0.067 = 0.7743

Thus,

P( 20 < x < 30) = 0.7743

So, we can conclude that the probability that a person getting married for the first time is in his or her twenties is 0.7743

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Answer:

a) P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33 and that represent the 33%

b) p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66

c) L_s =\frac{20}{30-20}=\frac{20}{10}=2 people

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Step-by-step explanation:

Notation

P represent the probability that the employee is idle

p_x represent the probability that the employee is busy

L_s represent the average number of people receiving and waiting to receive some information

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W_s represent the average time a person seeking information spends in the system

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This an special case of Single channel model

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Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}

And in order to find the probability we can do this:

P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33 and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

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And replacing we got:

L_s =\frac{20}{30-20}=\frac{20}{10}=2 people

Part d

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For the number of people wiating we can us ethe following formula"

L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}

And replacing we got this:

L_q =\frac{20^2}{30(30-20)}=1.333 people

Part e

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For this average we can use the following formula:

W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours

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