I don't know but I guess there is an answer
Answer:
(5,1) is the mid point of RS
1) Call x the sample mean = 3.56
2) Call s the sample standard deviation = 0.2
3) Given that the variable is normally distributed and the sample is large, you determine the interval of confidence from:
x +/- Z(0.5) s/√n
Wehre Z(0.5) is the value of the probabilities over 5% (90% of confidence mean to subtract 10%, which is 5% for each side (tails) of the normal distribuition) and is taken from tables.
Z(0.5) = 0.3085
Then the inteval is
x +/- 0.385 *s /√n = 3.56 +/- 0.385 * 0.2/√45
3.56 +/- 0.011 = ( 3.549, 3.571). This is the answer.
