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 1500 chocolate chips?(round to four decimal places as needed.)(b) the probability that a randomly selected bag contains fewer than 1050 chocolate chips is(round to four decimal places as needed.)(c) the probability that a randomly selected bag contains more than 1200 chocolate chips is(round to four decimal places as needed.)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 1252\) and \(\sigma=129\).

(a)

For \(x_1 = 1100\) and \(x_2 = 1500\):
\(z_1=\frac{1100 - 1252}{129}=\frac{- 152}{129}\approx - 1.18\)
\(z_2=\frac{1500 - 1252}{129}=\frac{248}{129}\approx1.92\)
Using the standard normal table \(P(-1.18<Z<1.92)=\Phi(1.92)-\Phi(-1.18)\)
\(\Phi(1.92)=0.9726\), \(\Phi(-1.18)=1 - \Phi(1.18)=1 - 0.8810 = 0.1190\)
\(P(-1.18<Z<1.92)=0.9726-0.1190 = 0.8536\)

(b)

For \(x = 1050\):
\(z=\frac{1050 - 1252}{129}=\frac{-202}{129}\approx - 1.57\)
Using the standard normal table \(P(Z < - 1.57)=1-\Phi(1.57)\)
\(\Phi(1.57)=0.9418\), so \(P(Z < - 1.57)=1 - 0.9418=0.0582\)

(c)

For \(x = 1200\):
\(z=\frac{1200 - 1252}{129}=\frac{-52}{129}\approx - 0.40\)
Using the standard normal table \(P(Z> - 0.40)=1-\Phi(-0.40)\)
Since \(\Phi(-0.40)=1 - \Phi(0.40)\) and \(\Phi(0.40)=0.6554\), then \(\Phi(-0.40)=0.3446\)
\(P(Z> - 0.40)=1 - 0.3446 = 0.6554\)

Answer:

(a) \(0.8536\)
(b) \(0.0582\)
(c) \(0.6554\)