Question

Help me with the following problem :(
Problem
Suppose that the sugar content for each orange is normally distributed with means of 0.5 and deviation of 0.05. What is the probability that a randomly selected orange has a sugar content between 0.54 and 0.61?

Answer 1

Solution :

Given that ,

mean = = 0.5

standard deviation = = 0.05  

P(0.54< x <0.61 ) = P[(0.54 - 0.5) /0.05 < (x - ) / < (0.61-0.5) /0.05 )]

= P(0.8 < Z <2.2 )

= P(Z <2.2 ) - P(Z < 0.8)

Using z table   

= 0.9861-0.7881

probability= 0.1980