Question

the number of chocolate chips in an 18 oz bag of chocolate chip cookies is approximately normally distributed with 1270 in the standard deviation is 130 chips

what is the probability that a randomly-selected bag contains no more than 1200 chips

what is the probability that a randomly-selected bag of 30 bags has a random mean that's no more than 1200 chips

what number of chocolate chips in a bag will correspond to the 15th percentile

Answer 1

Solution :

Given that ,

1) P(x < 1200)

= P[(x - ) / < (1200 - 1270) / 130]

= P(z < - 0.54)

Using z table,

= 0.2946

2) = 1270

= / n = 130 / 30 = 23.73

P( < 1200) = P(( - ) / < (1200 - 1270) / 23.73 )

= P(z < - 2.95)

Using z table

= 0.0016

Using standard normal table,

3) P(Z < z) = 15%

= P(Z < z ) = 0.15

= P(Z < - 1.04 ) = 0.15  

z = - 1.04

Using z-score formula,

x = z * +

x = - 1.04 * 130 + 1270

x = 1134.8