The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and a standard deviation of 117 chips. (a) Determine the 28th percentile for the number of chocolate chips in a bag. (b) Determine the number of chocolate chips in a bag that make up the middle 97% of bags. (c) What is the interquartile range of the number of chocolate chips in a bag of chocolate

Using z score formula:

X = z ∂ + µ

 = 157.833

Step-by-step explanation:

Solution:

Mean = µ = 1262

Standard deviation = ∂ = 117

(a)28th percentile for the number of chocolate chip.

P( z < z) = 28%

             = 0.28

P( z<- 0.58)  = 0.28

Z = -0.58

By using z score formula:

Z = x - µ /∂

-0.58= x – 117 / 1262

X = (- 0.58)(117) + (1262)

 =  1194.14

(b)Middle 97% of bag.

P(-z < z < z) = 97%

                    = 0.97

P( z < z) – p(z < -z) = 0.97

2p(z < z) -1 = 0.97

2p (z < z) = 1 + 0.97

P(z < z) = 1.97 / 2

            = 0.99

P(z < 2.33) = 0.99

Z ± 2.33

By using z score formula:

Z = x - µ / ∂

X = z ∂ + µ

  = - 2.33 x 117 + 1262

  =989.39

Z = 2.33

X = z ∂ + µ

  =  2.33 x 117 + 1262

 =1533.61

(c)What is the interquartile range of the number of chocolate chips in a bag of chocolate.

By using standard normal table,

The z dist’n formula:

P(z < z ) = 25%

            =0.25

P(z < -0.6745) = 0.25

Z = 0.6745

Using z score formula:

X = z∂ + µ

   = - 0.6745 x 117 + 1262

  = 1183.0835

First quartile = Q1 =1183.0835

The third quartile is:

P(z<z) = 75%

    = 0.75

P(z < 0.6745) = 0.75

Z = 0.6745

Using z score formula:

X = z ∂ + µ

  = 0.6745 x 117 + 1262

= 1340.9165

IQR = Q3 – Q1

     = 1340.9165 – 1183.0835

  = 157.833

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