Answer to Question #116594 in Statistics and Probability for sussana dzigbordi yevu

Question

1. Bags of smokeless fuel are labeled as weighing 100kg, but actual weights are normally distributed with a mean of 102kg and a standard deviation of 4kg.
(i) What proportion of bags will weigh more than 106kg? ii) What proportion of bags will weigh less than 95kg?
(iii) What proportion of bags will weigh between 91kg and 112kg?

Answer 1

(i) $P(X>106)=P(Z>\frac{106-102}{4}=P(Z>1)=1-P(Z<1)=0.1587.$

(ii) $P(X<95)=P(Z<\frac{95-102}{4})=P(Z<-1.75)=0.0401.$

(iii) $P(91<X<112)=P(\frac{91-102}{4}<Z<\frac{112-102}{4})=P(-2.75<Z<2.5)=$

$=P(Z<2.5)-P(Z<-2.75)=0.9908.$