Question

The amount of corn chips dispensed into a 13​-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 13.5 ounces and a standard deviation of 0.4 ounces. Suppose 50 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 50 bags exceeded 13.6 ounces. Round to four decimal places.

Answer 1

Solution :

Given that,

mean = = 13.5

standard deviation = = 0.4

n = 50

= = 13.5

= / n = 0.4 / 50 = 0.0566

P( > 13.6) = 1 - P( < 13.6)

= 1 - P[( - ) / < (13.6 - 13.5) /0.0566 ]

= 1 - P(z < 1.77)

Using z table,    

= 1 - 0.9616

= 0.0384