Statcrunchthe number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1261 chips and a standard deviation of 117 chips. (a) 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 97% of bags. (c) what is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?

A) First, we convert from percentiles to z-score using a z-table or graphing calculator. The z-table tells us that a z-score of about -0.64 is at the 26th percentile. Then, we convert from z-score to chips using the formula z = (x - mean)/standard deviation. -0.64 = (x - 1261)/117, so x = 1186.12, or about 1186 chips.
b) To find the percentage distance from the mean to one side of the distribution, we divide 97 by 2 to get 48.5. This means that 48.5% above and below the mean is the same as the middle 97%. To find the number of chocolate chips in the bag, we have to find number of chocolate chips in the 98.5th percentile (98.5 is found by adding 48.5 to 50) and the number of chocolate chips in the 1.5th percentile (1.5 is found by subtracting 48.5 from 50). We use a z-table to see that a z-score of about -2.17 is at the 1.5th percentile and a z-score of about 2.17 is at the 98.5th percentile. We convert -2.17 to chips using -2.17 = (x - 1261)/117, and x = 1007.11, or 1007 chips. We convert 2.17 to chips using 2.17 = (x - 1261)/117, and x = 1514.89, or 1515 chips. So a bag containing 1007 to 1515 chips makes the middle 97% of bags.
c) This question is similar to the previous question because it is basically asking you for the middle 50% of bags. The main difference is that we have to subtract the two values in this question to get one number. We divide 50 by 2 to find that 25% of the data falls above and below the mean. So, we need to find the value at the 25th percentile (50-25) and the 75th percentile (50+25). The z-table tells us that a z-score of about -0.67 is at the 25th percentile, and a z-score of about 0.67 is at the 75th percentile. Using the z-score formula, we find that a z-score of -0.67 is equivalent to 1,182.61 and a z-score of 0.67 is equivalent to 1339.39. The interquartile range = 1339.39-1182.61 = 156.78.

Ithink it would be 160 people