a volunteer ambulance service handles zero to five servicecalss on any given day

Question

a volunteer ambulance service handles zero to five servicecalss on any given day. The probabilty distribution for thenumber of calls follows No. of calls(X)                                0          1        2          3         4          5      P(X)                                               0.10    0.15    0.30     0.20     0.15      0.10 find the mean, variance and standard deviation for thedistribution a volunteer ambulance service handles zero to five servicecalss on any given day. The probabilty distribution for thenumber of calls follows No. of calls(X)                                0          1        2          3         4          5      P(X)                                               0.10    0.15    0.30     0.20     0.15      0.10 find the mean, variance and standard deviation for thedistribution P(X)                                               0.10    0.15    0.30     0.20     0.15      0.10 find the mean, variance and standard deviation for thedistribution

Explanation / Answer

mean= E(X)=0*0.1+1*0.15+2*0.3+3*0.2+4*0.15+5*0.1=2.45 E(X^2)=0*0.1+1^2*0.15+2^2*0.3+3^2*0.2+4^2*0.15+5^2*0.1=8.05 variance= E(X^2) - [E(X)]^2 =8.05 -2.45^2 =2.0475 standard deviation = variance =2.0475 =1.430909

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