Question

the following questions are either true or false.

1. Suppose that x is a normally distributed random variable, with a mean of 10 and a cariance of 4. then probability that x lies between 11.0 and 13.6 is 0.02726

2. Find E(X)= mean id X is a normally distributed random vaeiqvle with standard deviation 1 and if X is a normally distributed random variable with standard deviation 1 and with P(x>5)=0.6915 mean= E(X) =5.50

3. Find P(-0.47<Z<0.94)=0.5666

4. In a random sample of 10,000 claims filed against an automobile insurance company, 75% exceed $300. What is the probability that of the next 400 claims filed more than 72% will be above 300$? Therefore, the probability that there are more than 288 claims of 300$ is 0.95.

Answer 1

1)

probability =P(11<X<13.6)=P((11-10)/2)<Z<(13.6-10)/2)=P(0.5<Z<1.8)=0.9641-0.6915=0.2726

True

2)

probability =P(X>5)=P(Z>(5-5.5)/1)=P(Z>-0.5)=1-P(Z<-0.5)=1-0.3085=0.6915

True

3)

probability =P(-0.47<X<0.94)=P((-0.47-0)/1)<Z<(0.94-0)/1)=P(-0.47<Z<0.94)=0.8264-0.3192=0.5072

False

4)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.7500
sample size       =n=	400
std error of proportion=σp=√(p*(1-p)/n)=	0.0217

probability =P(X>0.72)=P(Z>(0.72-0.75)/0.022)=P(Z>-1.39)=1-P(Z<-1.39)=1-0.0823=0.9177

False (since above is not equal to 0.95)