The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1261 and a standard deviation of 118. (a) Determine the 30th 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 98% of bags. (c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?

(A) 1199.168

(B) 1503.372

(C) 159.17728

Step-by-step explanation:

(A) To determine the 30th percentile for the number of chocolate chips in the bag, we find the z-score for the 30th percentile.

Found using a z-table or z-calculator, the z-score for the 30th percentile is -0.524

The formula for finding X (the number of items in a given percentile) is:

X = M + Z(S.D.)

Where M is the mean, Z is the specific z-score of the sought percentile and S.D. is the standard deviation.

So for the 30th percentile,

X = 1261 + (-0.524)(118)

X = 1261 - 61.832 = 1199.168

(B) The number of chocolate chips that make up the middle 98% of chips in the bag is

X = 1261 + (2.054)(118)

X = 1261 + 242.372 = 1503.372

(C) For normal distributions, Interquartile range is Q3 - Q1, that is; 3rd quartile minus 1st quartile.

This is within 1.34896 standard deviations of the mean.

IQR = (1.34896)(118)

IQR = 159.17728

hence, the required result is

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

we have to compute b-a which means subtracting corresponding elements of a from b.

-

any of them that add up to 180 degrees.