The weight of a bag of corn chips is normally distributed wi

The weight of a bag of corn chips is normally distributed with a mean of 22 ounces and a standard deviation of

mean = 22 oz

sd = 0.5 oz

x1 = 21.2

z1 = (x-mean)/sd = (21.2-22)/0.8 = -1.6

P(z<-1.6) = 0.0548

so, 5.48%

x2 = 21.7

z2 = (21.7-22)/0.5 = -0.6

P(z>-0.6) = 1 -P(z<-0.6) = 1 - 0.2743 = 0.7257

x3= x3

z = z3

P(z>z3) = 0.2

P(z<z3) = 1 - P(z>z3) = 1 - 0.2 = 0.8

P(z<z3) = 0.8

z3 = 0.84

x3 = 22 + 0.84*0.5 = 22.42 oz

x4=21 oz

z4 = (21-22)/0.5 = -2

P(z<-2) = 0.0228

so, 2.28% of bag weight less than 21 oz

so, it not accepted as per FDA standards because it limits to 1% only.

x5 =21

now to minimize dispersion, new standard deviation to be applied,

so sd=sd2

z5 = (21-22)/sd2 = -1/sd

P(z<z5) = 0.01

z5 = -3.09

-1/sd2 = -3.09

sd2 = 1/3.09

sd2 = 0.3236

in order to maintain FDA norms, dispersion to be minimised to 0.3236 oz.