a)
for normal distribution z score =(X-μ)/σ | |
here mean= μ= | 100 |
std deviation =σ= | 30.0000 |
P(a student take longer than an hour to complete this forum:
probability = | P(X>60) | = | P(Z>-1.333)= | 1-P(Z<-1.33)= | 1-0.0918= | 0.9082 |
hence P(exactly 2 out of 4 will take longer than 1 hour to complete)=4C2(0.9082)2(0.0918)2 =0.0417
b)as range 130-150 falls at the maximum distance , therefore it is least likely to lie in range 130-150
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