Question

Let X be the weight of a randomly selected 10oz bag of chips. Suppose that X has a normal distribution with a mean of 10.2 and standard deviation of .05. Find the weight of x* so that 95% of all 10oz bags have a weight of at least x*.

Answer 1

Solution :

mean = =  10.2

standard deviation = =0.05

n = 10

= 10.2

  =  ( /n) = (0.05/ 10 ) = 0.0158

Using standard normal table,

P(Z > z) = 95%

1 - P(Z < z) = 0.95

P(Z < z) = 1 - 0.95= 0.05

P(Z <-1.645) = 0.05

z = - 1.65

Using z-score formula,

= z * .  +

= -1.65 *  0.0158 + 10.2

= 10.17

X^* = 10.17