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.
(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 1000 chocolate chips?
(c) what proportion of bags contains more than 1200 chocolate chips?
(d) what is the percentile rank of a bag that contains 1475 chocolate chips?
(a) the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips, inclusive, is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1252\), \(\sigma=129\).
For \(x = 1100\), \(z_1=\frac{1100 - 1252}{129}=\frac{- 152}{129}\approx - 1.18\)
For \(x = 1400\), \(z_2=\frac{1400 - 1252}{129}=\frac{148}{129}\approx1.15\)

Step2: Find the probabilities

Using the standard normal distribution table or a calculator with a normal - distribution function (\(P(Z < z)\)).
\(P(Z<-1.18) = 0.1190\), \(P(Z < 1.15)=0.8749\)

Step3: Calculate the probability between the two z - scores

\(P(-1.18\leq Z\leq1.15)=P(Z < 1.15)-P(Z<-1.18)\)
\(P(-1.18\leq Z\leq1.15)=0.8749 - 0.1190=0.7559\)

Answer:

\(0.7559\)