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Yuri [45]
3 years ago
15

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)
Mathematics
2 answers:
wel3 years ago
5 0

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

MA_775_DIABLO [31]3 years ago
3 0

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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<h2>what these three shapes have in common is that they are Parallelograms.</h2>

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3. Consecutive angles are supplementary (<u>angles that add up to 180°</u>)

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Now, the <u>differences between the Rectangle and the Rhombus</u> are:

1. In the Rectangle only the opposite sides are of the same length, while in the rhombus all of its sides are of the same length.

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3 years ago
Which of the following side lengths can form a triangle?
Margaret [11]

Okay, so my last answer was deleted because I didnt show any work.


According to the rule,


a+b>c

c+b>a

c+a>b


so...


2+3=6...6<7

21+23=44...44+44

12+36=48...48<53

14+17=31...31>30


Since answer D is the only answer that fullfills the rule a+b>c, it is the correct answer


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Simora [160]

(a) From the histogram, you can see that there are 2 students with scores between 50 and 60; 3 between 60 and 70; 7 between 70 and 80; 9 between 80 and 90; and 1 between 90 and 100. So there are a total of 2 + 3 + 7 + 9 + 1 = 22 students.

(b) This is entirely up to whoever constructed the histogram to begin with... It's ambiguous as to which of the groups contains students with a score of exactly 60 - are they placed in the 50-60 group, or in the 60-70 group?

On the other hand, if a student gets a score of 100, then they would certainly be put in the 90-100 group. So for the sake of consistency, you should probably assume that the groups are assigned as follows:

50 ≤ score ≤ 60   ==>   50-60

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90 < score ≤ 100   ==>   90-100

Then a student who scored a 60 should be added to the 50-60 category.

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