Spring2018-Math 1040-40-Online-Livu Homework: 7.2 Homework Score: 0 of 1 pt 7.2.

Question

Spring2018-Math 1040-40-Online-Livu Homework: 7.2 Homework Score: 0 of 1 pt 7.2.49 20of22 (1 8 complete) The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean 1263 and a standard deviation of 118 (a) Determine the 27th percente for the number of chocolate chipsabag (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 chooolate chips in a bag of chocolate chip cookies? Click here to view the standard nermal distribution table.age (a) The 27m percortile for the number of chocolate chips in a bag ofchocolato chip cookies ischocolate chps (Round to the nearest whole number as needed.) Enter your answer in the answer box and then click Check Answer 2 patanng Clear Al 3 4 6

Explanation / Answer

Solution:- Given that mean = 1263 , sd = 118

a) 27th percentile implies P(Z < z) = 0.2700
z value which separates the lower 0.2700 area from the rest is - 0.61
Formula: z = (X-Mean)/SD
X = Mean + (z*SD)
X = 1263 + (- 0.6128*118)
X = 1190.6896
X = 1191 ( nearest number)

=> The 27th percentile for the nuumber of chocolate chips in a big of chocolate chip cookies is 1191 chocolate chips

b) Middle 98% leaves 2% area equally in the left and right tails of the normal curve
Area in the left tail = Area in the right tail = 2%/2 = 1% = 0.01
The z value which separates the bottom 0.01 area from the rest is - 2.33 approximately
The z value which separates the top 0.01 area from the rest is + 2.33 approximately
X1 = 1263 + (-2.33*118) = 988.06 = 988
X2 = 1263 + (2.33*118) = 1537.94 = 1538
Required interval of the chocolate chips is from 988 to 1538

c)

Q1 = 1263 +(-0.6745*118) = 1183.409 = 1183

Q3 = 1263 +(0.6745*118) = 1342.591 = 1343

IQR = Q3 - Q1 = 1343 - 1183 = 160

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