Question

The number of chocolate chips in an​ 18-ounce bag of chocolate chip cookies is approximately normally distributed with a mean of 1252 chips and standard deviation 129 chips.

​(b) What is the probability that a randomly selected bag contains fewer than 1100 chocolate​ chips?

​(c) What proportion of bags contains more than 1175 chocolate​ chips?

​(d) What is the percentile rank of a bag that contains 1425 chocolate​ chips?

Answer 1

solution

(A)P(X<1100 ) = P[(X- ) / < (1100-1252) / 129]

= P(z <-1.18 )

Using z table

= 0.1190

probbaility=0.1190

(B)

P(x >1175 ) = 1 - P(x<1175 )

= 1 - P[(x -) / < (1175-1252) / 129 ]

= 1 - P(z <-0.60 )

Using z table

= 1 -  0.2743

=0.7257

proportion=0.7257

(C)

P(X<1425 ) = P[(X- ) / < (1425-1252) / 129]

= P(z <1.34 )

Using z table

= 0.9099

answer=90.99%