Standard Deviation

Question

Weights of packages of M&M candies are Normally distributed with the average weight of 0.91g and the standard deviation of 0.15g. If a random sample of 32 packages is selected, what is the probability that their average weight is between 0.87g and 0.94g?

Answer 1

  z = (mean - sample mean)/ ( stddev/sqrt(N))

 

 stddev/sqrt(n) = 0.15/sqrt(32) = 0.0265165

 

Prob

 (0.87 - 0.91)/0.0265165 < Z < (0.94 - 0.91)/0.0265165

 

= Prob ( -1.508494711 < Z < 1.1313711)

 

= Prob ( z < 1.1313711) - Prob( z < -1.508494711)

 

 = 0.8708 - 0.0655 

 

= 0.8053