Use uniform disrtribution-A particular employee arrives to work some time betwee

Question

Use uniform disrtribution-A particular employee arrives to work some time between 8:00 am - 8:30 am. Based on past experience the Company has determined that the employee is equally likely to arrive at any time between 8:00 am - 8:30 am.use uniform distribution..A) On average, what time does the employee arrive?b) What is the standard deviation of the time at which the employee arrives? c)if a call comes in for the employee at 8.10 am find the probability that the employee will be there to take the call d)Find the probability that the employee arrives between 8:20 eam - 8:25 am? e)find the probability that the employee will arrive after 8.15 am.F)find the probability that the employee will arrive exactly at 8.10 am.

Explanation / Answer

Let

x = 0 be 8:00 and
x = 30 be 8:30

a)

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    30      
          
Thus, the mean,      
          
u = mean = (b + a)/2 =    15

Hence, the average arrival time is 8:15 am. [ANSWER]
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b)

Note that here,  
  
a = lower fence of the distribution =    0
b = upper fence of the distribution =    30
  
Thus, the standard deviation is  
  
s^2 = variance = (b -a)^2 / 12 =    75

s = standard deviation = sqrt(s^2) =    8.660254038 minutes [ANSWER]

*************************

c)

Hence, x must be less than 10.

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    30      
          
Note that P(x<c) = P(a<x<c) = (c-a)/(b-a). Thus, as          
          
c = critical value =    10      
          
Then          
          
P(x<c) =    0.333333333   [ANSWER]  

*************************

d)

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    30      
          
Thus, the mean, variance, and standard deviations are          
          
u = mean = (b + a)/2 =    15      
s^2 = variance = (b -a)^2 / 12 =    75      
s = standard deviation = sqrt(s^2) =    8.660254038      
          
Thus, the area between the said numbers is          
          
c = lower number =    20      
d = higher number =    25      
          
Thus, the probability between these two values is          
          
P = (d - c)/(b - a) =    0.166666667   [ANSWER]
*******************************

e)

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    30      
          
          
Note that P(x>c) = P(c<x<b) = (b-c)/(b-a). Thus, as          
          
c = critical value =    15      
          
Then          
          
P(x>c) =    0.5   [ANSWER]

****************************

f)

We need an interval for probabilities. The probability of exact point times is

P(x = 10) = 0 [ANSWER]

*****************************

For f, if you meant 8:10 as in 8:10:00 up to just before 8:11, then it would be:

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    30      
          
Thus, the mean, variance, and standard deviations are          
          
u = mean = (b + a)/2 =    15      
s^2 = variance = (b -a)^2 / 12 =    75      
s = standard deviation = sqrt(s^2) =    8.660254038      
          
Thus, the area between the said numbers is          
          
c = lower number =    10      
d = higher number =    11      
          
Thus, the probability between these two values is          
          
P = (d - c)/(b - a) =    0.033333333   [ALTERNATE ANSWER FOR F]  

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