Asked by Gayle
                The number of ships to arrive at a harbor on any given day is a random variable represented by x. The probability distribution of x is as follows. (Give your answers correct to two decimal places.) 
x 10 11 12 13 14
P(x) 0.37 0.09 0.05 0.14 0.35
(a) Find the mean of the number of ships that arrive at a harbor on a given day.
  
(b) Find the standard deviation, ó, of the number of ships that arrive at a harbor on a given day.
            
            
        x 10 11 12 13 14
P(x) 0.37 0.09 0.05 0.14 0.35
(a) Find the mean of the number of ships that arrive at a harbor on a given day.
(b) Find the standard deviation, ó, of the number of ships that arrive at a harbor on a given day.
Answers
                    Answered by
            Kuai
            
    mean = x*p(x)
a) 10(.37) + 11(.09) + 12(.05) + 13(.14) + 14(.35)
3.7 + 0.99 + 0.6 + 1.82 + 4.9 = 12.01
x^2*p(x)
variance = 10^2 (.37) + 11^2 (0.99) + 12^2 (0.05) +13 ^2 (0.14) +14^2 (0.35)
37 + 119.79 + 7.2 + 23.66 + 68.6 = 256.25
Standard deviation = sqrt(variance - xbar^2)
=sqrt(256.25 - 12.01^2))
= sqrt(256.25- 144.2401)
= sqrt(112.0099)
= 10.58
       
    
a) 10(.37) + 11(.09) + 12(.05) + 13(.14) + 14(.35)
3.7 + 0.99 + 0.6 + 1.82 + 4.9 = 12.01
x^2*p(x)
variance = 10^2 (.37) + 11^2 (0.99) + 12^2 (0.05) +13 ^2 (0.14) +14^2 (0.35)
37 + 119.79 + 7.2 + 23.66 + 68.6 = 256.25
Standard deviation = sqrt(variance - xbar^2)
=sqrt(256.25 - 12.01^2))
= sqrt(256.25- 144.2401)
= sqrt(112.0099)
= 10.58
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