Question

he 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.

​(a) What is the probability that a randomly selected bag contains between 1100 and 1400 chocolate​ chips, inclusive?

​(b) What is the probability that a randomly selected bag contains fewer than 1025 chocolate​ chips?

​(c) What proportion of bags contains more than 1200 chocolate​ chips?

​(d) What is the percentile rank of a bag that contains 1050 chocolate​ chips?

Answer 1

Solution :

Given that ,

mean = = 1252

standard deviation = = 129

a) P(1100 < x < 1400) = P[(1100 - 1252) / 129) < (x - ) /  < (1400 - 1252) / 129) ]

= P(-1.18 < z < 1.15)

= P(z < 1.15) - P(z < -1.18)

Using z table,

= 0.8749 - 0.119

= 0.7559

b) P(x < 1025) = P[(x - ) / < (1025 - 1252) /129 ]

= P(z < -1.76)

Using z table,

= 0.0392

c) P(x > 1200) = 1 - p( x< 1200)

=1- p P[(x - ) / < (1200 - 1252) / 129]

=1- P(z < -0.40)

Using z table,

= 1 - 0.3446

= 0.6554

d) x = 1050

Using z-score formula,

z = x - /   

z = 1050 - 1252 / 129

z = -1.57

= P(z < -1.57)

Using z table,

= 0.0582

The percentage is = 5.82%

percentile rank is = 6th