Question 1184737
Since we are given that the weights are normally distributed with {{{mu = 491}}} and {{{sigma = 9}}}, 

we want to solve for {{{alpha}}} in the probability equation {{{P("Xbar" > alpha) = 0.03}}}.


===> {{{P((Xbar - mu)/(sigma/sqrt(n))= Z > (alpha-mu)/(sigma/sqrt(n))=  (alpha-491)/(9/sqrt(16)) ) =  0.03)}}}.


Now {{{P(Z > 1.881) = 0.03}}}


===> {{{(alpha-491)/(9/sqrt(16)) = 1.881}}}  ====> {{{alpha = 495.23225}}}, or 495 grams to the nearest gram.


Therefore, if you pick 16 fruits at random, then 3% of the time, their mean weight will be greater than {{{red(495)}}} grams.


Note that, even though the sample size is < 30, we know by the CLT that the sampling distribution of the mean follows almost 
a normal distribution whose parameters are known from the underlying population.


Probabilities were taken from https://stattrek.com/online-calculator/normal.aspx.