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 1000 and 1500 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 1425 chocolate​ chips?

Answer 1

Answer:

(a) P(1000 < X < 1500) = 0.0256

(b) P(X < 1025) = 0.0392

(c) P(X > 1200) = 0.6554

(d) Percentile rank of a bag that contains 1425 chocolate​ chips = 90.98%

Step-by-step explanation:

We are given that 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.

Firstly, Let X = number of chocolate chips in a bag

The z score probability distribution for is given by;

         Z = $\frac{ X - \mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean = 1252 chips

           $\sigma$ = standard deviation = 129 chips

(a) Probability that a randomly selected bag contains between 1000 and 1500 chocolate​ chips, inclusive is given by = P(1000 $\leq$ X $\leq$ 1500) = P(X $\leq$ 1500) - P(X < 1000)

 P(X $\leq$ 1500) = P( $\frac{ X - \mu}{\sigma}$ $\leq$ $\frac{1500-1252}{129}$ ) = P(Z $\leq$ 1.92) = 0.9726

 P(X < 1000) = P( $\frac{ X - \mu}{\sigma}$ < $\frac{1000-1252}{129}$ ) = P(Z < -1.95) = 1 - P(Z $\leq$ 1.95)

                                                       = 1 - 0.9744 = 0.0256

Therefore, P(1000 $\leq$ X $\leq$ 1500) = 0.9726 - 0.0256 = 0.947

(b) Probability that a randomly selected bag contains fewer than 1025 chocolate​ chips is given by = P(X < 1025)

   P(X < 1025) = P( $\frac{ X - \mu}{\sigma}$ < $\frac{1025-1252}{129}$ ) = P(Z < -1.76) = 1 - P(Z $\leq$ 1.76)

                                                          = 1 - 0.9608 = 0.0392

(c) Proportion of bags contains more than 1200 chocolate​ chips is given by = P(X > 1200)

    P(X > 1025) = P( $\frac{ X - \mu}{\sigma}$ > $\frac{1200-1252}{129}$ ) = P(Z > -0.40) = P(Z < 0.40) = 0.6554

(d) Percentile rank of a bag that contains 1425 chocolate​ chips is given by;

Firstly we will calculate the z score of 1425 chocolate chips, i.e.;

                Z = $\frac{1425-1252}{129}$ = 1.34

Now, we will check the area probability in z table which corresponds to this critical value of x;

The value which we get is 0.9098.

Therefore, 90.98% is the rank of bag that contains 1425 chocolate​ chips.