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 0.7559.
(round to four decimal places as needed.)
(b) the probability that a randomly selected bag contains fewer than 1000 chocolate chips is 0.0256.
(round to four decimal places as needed.)
(c) the proportion of bags that contains more than 1200 chocolate chips is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1252\) (mean) and \(\sigma=129\) (standard deviation). For \(x = 1200\), \(z=\frac{1200 - 1252}{129}=\frac{-52}{129}\approx - 0.4031\)

Step2: Find the probability

We want \(P(X>1200)\). Since \(P(X>1200)=1 - P(X\leq1200)\). Using the standard normal distribution table or a calculator with a normal - distribution function (e.g., in Excel: NORM.S.DIST(-0.4031,TRUE)), \(P(Z\leq - 0.4031)\approx0.3436\)

Step3: Calculate the final probability

\(P(X>1200)=1 - 0.3436 = 0.6564\)

Answer:

\(0.6564\)