The number of chocolate chips in a popular brand of cookie is normally distributed with a mean of 22 chocolate chips per cookie and a standa

Question

The number of chocolate chips in a popular brand of cookie is normally distributed with a mean of 22 chocolate chips per cookie and a standard deviation of 3.5 chips. When the cookies come out of the oven, only the middle 90% in terms of the number of chocolate chips are acceptable (the rest are considered defective). What are the cutoff numbers for the number of chocolate chips in acceptable cookies? (Give your answers to three decimal places)

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Kylie 2 weeks 2021-09-15T02:58:43+00:00 2 Answers 0

Answers ( )

    0
    2021-09-15T02:59:47+00:00

    Answer:

    16,242 and 27,758

    Step-by-step explanation:

    The first thing to keep in mind is that

    middle 90% means 5% bottom and 5% top are excluded.

    we have that the mean (m) is 22 and the standard deviation (sd) 3,5.

    using percentile table, we will note down the z score corresponding to 5th and 95th percentile

    z (5th) = -1.645

    z (95th) = 1,645

    lower cut off value =

    m + z (5th) * sd

    22 + (-1,645) * 3,5

    = 16,242

    upper cut value

    m + z (95th) * sd

    22 + (1,645) * 3.5

    = 27,758

    Therefore the values are 16,242 and 27,758

    0
    2021-09-15T03:00:07+00:00

    Answer:

    The cutoff numbers for the number of chocolate chips in acceptable cookies are 16.242 and 27.758

    Step-by-step explanation:

    Problems of normally distributed samples can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    In this problem, we have that:

    \mu = 22, \sigma = 3.5

    Middle 90%

    Between the 50 – (90/2) = 5th percentile to the 50 + (90/2) = 95th percentile.

    5th percentile:

    X when Z has a pvalue of 0.05. So X when Z = -1.645.

    Z = \frac{X - \mu}{\sigma}

    -1.645 = \frac{X - 22}{3.5}

    X - 22 = -1.645*3.5

    X = 16.242

    95th percentile:

    X when Z has a pvalue of 0.95. So X when Z = 1.645.

    Z = \frac{X - \mu}{\sigma}

    1.645 = \frac{X - 22}{3.5}

    X - 22 = 1.645*3.5

    X = 27.758

    The cutoff numbers for the number of chocolate chips in acceptable cookies are 16.242 and 27.758

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