Respuesta :
Answer:
Step-by-step explanation:
here mean difference =130-120 =10
and standard deviation of mean difference =(8^2/3+10^2/3)=7.3937
therefore from normal distribution; probability that mean difference is higher than 5:
P(X>5)=P(Z>(5 - 10)/7.3937)
=P(Z>-0.68)
=0.7517
(please try 0.7506 if this comes wrong due to rounding of z score)
Answer:
The probability that the mean IQ of the 3 North Catalina students is at least 5 points higher than the mean IQ of the 3 Chapel Mountain students = 0.7517
Step-by-step explanation:
let A =X-Y where X and Y are IQ of 3 North Catalina student and mean IQ of the 3 Chapel Mountain students
Hence expected value of A =130-120 =10
std error of mean difference =(82/3+102/3)1/2 =7.3937
Hence probability that the mean IQ of the 3 North Catalina students is at least 5 points higher than the mean IQ of the 3 Chapel Mountain students =P(A>5)=P(Z>(5-10)/7.3937)=P(Z>-0.68)=0.7517