a)
for normal distribution z score =(X-μ)/σx | |
here mean= μ= | 2 |
std deviation =σ= | 1.000 |
probability =P(X<2.6)=(Z<(2.6-2)/1)=P(Z<0.6)=0.7257 |
b)
sample size =n= | 2 |
std error=σx̅=σ/√n= | 0.70711 |
probability =P(X>3)=P(Z>(3-2)/0.707)=P(Z>1.41)=1-P(Z<1.41)=1-0.9207=0.0793 |
c)
probability =P(X>2.7)=P(Z>(2.7-2)/0.707)=P(Z>0.99)=1-P(Z<0.99)=1-0.8389=0.1611 |
d)
sample size =n= | 100 |
std error=σx̅=σ/√n= | 0.10000 |
probability =P(1.85<X<16.1)=P((1.85-2)/0.1)<Z<(16.1-2)/0.1)=P(-1.5<Z<141)=1-0.0668=0.9332 |
Suppose that Xi are IID normal random variables with mean 2 and variance 1, for i...
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