Question

the number of chocolate chips in an 18 - ounce bag of chocolate chip cookies is approximately normally distributed with mean 1252 and standard deviation 129 chips. (a) what is the probability that a randomly selected bag contains between 1100 and 1400 chocolate chips? (round to four decimal places as needed.) (b) the probability that a randomly selected bag contains fewer than 1025 chocolate chips is (round to four decimal places as needed.) (c) the proportion of bags that contains more than 1175 chocolate chips is (round to four decimal places as needed.) (d) a bag that contains 1000 chocolate chips is in the □ rd percentile. (round to the nearest integer as needed.)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1252\) and \(\sigma=129\).
For part (a):
When \(x_1 = 1100\), \(z_1=\frac{1100 - 1252}{129}=\frac{- 152}{129}\approx - 1.18\)
When \(x_2 = 1400\), \(z_2=\frac{1400 - 1252}{129}=\frac{148}{129}\approx1.15\)
Using the standard normal table, \(P(-1.18\lt Z\lt1.15)=\Phi(1.15)-\Phi(-1.18)\)
\(\Phi(1.15) = 0.8749\), \(\Phi(-1.18)=0.1190\)
\(P(-1.18\lt Z\lt1.15)=0.8749 - 0.1190=0.7559\)

For part (b):
When \(x = 1025\), \(z=\frac{1025 - 1252}{129}=\frac{-227}{129}\approx - 1.76\)
Using the standard normal table, \(\Phi(-1.76)=0.0392\)

For part (c):
When \(x = 1175\), \(z=\frac{1175 - 1252}{129}=\frac{-77}{129}\approx - 0.60\)
\(P(X\gt1175)=1-\Phi(-0.60)\)
\(\Phi(-0.60) = 0.2743\), so \(P(X\gt1175)=1 - 0.2743=0.7257\)

For part (d):
When \(x = 1000\), \(z=\frac{1000 - 1252}{129}=\frac{-252}{129}\approx - 1.95\)
Using the standard normal table, \(\Phi(-1.95)=0.0256\), so the percentile is approximately \(3\)rd.

Answer:

(a) \(0.7559\)
(b) \(0.0392\)
(c) \(0.7257\)
(d) \(3\)