Mathematics, 14.06.2020 00:57 leannaadrian

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? (b) What is the probability that a randomly selected bag contains fewer than 1125 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?

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Answer from: calibaby1220

a) 85.36% probability that a randomly selected bag contains between 1100 and 1500 chocolate chips

b) 16.35% probability that a randomly selected bag contains fewer than 1125 chocolate chips

c) 0.3446 = 34.46% of bags contains more than 1200 chocolate chips

d) 91st percentile.

Step-by-step explanation:

Problems of normally distributed samples are 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 question, we have that:

$\mu = 1252, \sigma = 129$

(a) What is the probability that a randomly selected bag contains between 1100 and 1500 chocolate chips?

This is the pvalue of Z when X = 1500 subtracted by the pvalue of Z when X = 1100.

X = 1500

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

$Z = \frac{1500 - 1252}{129}$

Z = 1.92

Z = 1.92 has a pvalue of 0.9726

X = 1100

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

$Z = \frac{1100 - 1252}{129}$

Z = -1.18

Z = -1.18 has a pvalue of 0.1190

0.9726 - 0.1190 = 0.8536

85.36% probability that a randomly selected bag contains between 1100 and 1500 chocolate chips

(b) What is the probability that a randomly selected bag contains fewer than 1125 chocolate chips?

This is the pvalue of Z when X = 1125. So

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

$Z = \frac{1125 - 1252}{129}$

Z = -0.98

Z = -0.98 has a pvalue of 0.1635

16.35% probability that a randomly selected bag contains fewer than 1125 chocolate chips.

This is 1 subtracted by the pvalue of Z when X = 1200.

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

$Z = \frac{1200 - 1252}{129}$

Z = -0.40

Z = -0.40 has a pvalue of 0.3446

0.3446 = 34.46% of bags contains more than 1200 chocolate chips

(d) What is the percentile rank of a bag that contains 1425 chocolate chips?

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

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

Z = 1.34

Z = 1.34 has a pvalue of 0.91

So the answer is the 91st percentile.

Answer from: Quest

Best answer that i found on yahoo ok set : x= length of 2nd piece in cm x- 3= length of 1st piece in cm x-4= length of 3rd piece in cm then since you know they all came from the 20 cm ribbon you add them together and set them equal to 20x+ x-3 +x-4 = 20 3x - 7= 20 3x= 27 x= 9 the shortest piece is the third piece which is x- 4 x-4 9-4 = 5 the length of the smallest piece is 5 cm. the other lengths are 9 cm 6 cm

Answer from: Quest

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step-by-step explanation:

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