Answer:
a) 37.31 b) 42.70 c) 0.57 d) 0.09
Step-by-step explaanation:
We are regarding a normal distribution with a mean of 35 and a standard deviation of 6, i.e., 
= 35 and 
= 6. We know that the probability density function for a normal distribution with a mean of and a standard deviation of is given by
![f(x) = \frac{1}{\sqrt{2\pi}\sigma}\exp[-\frac{(x-\mu)^{2}}{2\sigma^{2}}]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%5Cpi%7D%5Csigma%7D%5Cexp%5B-%5Cfrac%7B%28x-%5Cmu%29%5E%7B2%7D%7D%7B2%5Csigma%5E%7B2%7D%7D%5D)
in this case we have
![f(x) = \frac{1}{\sqrt{2\pi}6}\exp[-\frac{(x-35)^{2}}{2(6^{2})}]](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B2%5Cpi%7D6%7D%5Cexp%5B-%5Cfrac%7B%28x-35%29%5E%7B2%7D%7D%7B2%286%5E%7B2%7D%29%7D%5D)
Let 
be the random variable that represents a row score, we find the values we are seeking in the following way
a) we are looking for a number 
such that
= 
, this number is =37.31
you can find this answer using the R statistical programming languange and the instruction qnorm(0.65, mean = 35, sd = 6)
b) we are looking for a number 
such that
= 
, this number is =42.70
you can find this answer using the R statistical programming languange and the instruction qnorm(0.9, mean = 35, sd = 6)
c) we find this probability as
=
you can find this answer using the R statistical programming languange and the instruction pnorm(38, mean = 35, sd = 6) -pnorm(28, mean = 35, sd = 6)
d) we find this probability as
=
you can find this answer using the R statistical programming languange and the instruction pnorm(44, mean = 35, sd = 6) -pnorm(41, mean = 35, sd = 6)
