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

step-by-step explanation:

the word superfluous means unnecessary

Answer from: Quest

Answer: question no. 1 -a. yes, the sides are in the ratio 2 : 5.question no. 2 -b. 59°question no. 3 -c. angles x and y are remote interior angles.question no. 4 -d. 90°, 45° and 45°question no. 5 -b. mn and qrstep-by-step explanation: question no. 1 -rs : xu = 3 : 7.5 = 2 : 5rt : x? = 6.4 : 16 = 2 : 5st : u? = 6 : 15 = 2 : 5as the ratios of all sides are equivalent, the triangles must be congruent and the sides must be proportional.therefore, the answer is: a. yes, the sides are in the ratio 2 : 5.question no. 2 -in the second diagram, angle b is the only letter not shown. therefore, the number must be equivalent to angle b.if we substitute the value of angle b as being 22° in the first diagram, we can now use the angle sum of a triangle to obtain the value of angle a.angle sum of a triangle = 180°therefore: let angle a = aa + 99 + 22 = 180a = 180 - 99 - 22a = 59°therefore, the answer is: b. 59°question no. 3 -angles x and y don't share the vertex of the triangle with the exterior angle w. remote interior angles don't share a vertex of a triangle with an exterior angle. this means that angles x and y are remote interior angles.therefore, the answer is: c. angles x and y are remote interior angles.question no. 4 -the sum of all interior angles of a triangle must be equivalent to 180°.90°, 45° and 45° are the only set of angle measures that are equivalent to 180°.therefore, the answer is: d. 90°, 45° and 45°question no. 5 -the only pair of sides stated in the options which are corresponding are mn and qr.therefore, the answer is: b. mn and qr.

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