Mathematics, 06.07.2021 07:40 ronniethefun

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a standard deviation of 118. Determine the 26th 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 95% of bags. (c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?

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Answer from: drcarver

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that $\mu = 1262, \sigma = 118$

a) Determine the 26th percentile for the number of chocolate chips in a bag.

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

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

$-0.643 = \frac{X - 1262}{118}$

X - 1262 = -0.643*118

X = 1186

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

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

$-1.96 = \frac{X - 1262}{118}$

X - 1262 = -1.96*118

X = 1031

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

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

$1.96 = \frac{X - 1262}{118}$

X - 1262 = 1.96*118

X = 1493

Between 1031 and 1493.

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 1262}{118}$

X - 1262 = -0.675*118

X = 1182

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 1262}{118}$

X - 1262 = 0.675*118

X = 1342

IQR:

1342 - 1182 = 160

Answer from: Quest

p> q if you calculate p is100 q is 79

What do you know to be true about the values of p and q?

Answer from: Quest

Where the white dots are is where your points should be plotted.a translation is simply just moving something from one place to another. in this case, you would move each plot point of the original shape 2 units up, and 5 units to the left.

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The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distribut...