The amount of cereal that can be poured into a small bowl varies with a mean of
ID: 3067069 • Letter: T
Question
The amount of cereal that can be poured into a small bowl varies with a mean of
1.31.3 ounces and a standard deviation of 0.3 ounce. A large bowl holds a mean of 2.2ounces with a standard deviation of 0.4 ounce. A student opens a new box of cereal and pours one large and one small bowl. Assume that the amounts poured into the two bowls are independent. Complete parts? (a) through? (f) below.
?a) How much more cereal does the student expect to be in the large? bowl?-------?ounce(s) ?(Type an integer or a? decimal.)
?b) What is the standard deviation of this? difference?----------?ounce(s)
?(Round to two decimal places as? needed.)
?c) If the difference follows a Normal? model, what is the probability the small bowl contains more cereal than the large? one?-------
?(Round to three decimal places as? needed.)
?d) What are the mean and standard deviation of the total amount of cereal in the two? bowls?
The mean is------?ounce(s). ?(Type an integer or a? decimal.)
The standard deviation is ------- ounce(s).
?(Round to two decimal places as? needed.)
?e) If the total follows a Normal? model, what is the probability the student poured out more than 4.1 ounces of cereal in the two bowls? together?------ ?(Round to three decimal places as? needed.)
?f) The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounce. Find the expected amount of cereal left in the box and the standard deviation. The expected amount of cereal left in the box is--------?ounce(s).
?(Type an integer or a? decimal.)
The standard deviation is -------- ounce(s). (Round to two decimal places as? needed.)
Explanation / Answer
a)How much more cereal does the student expect to be in the large? bowl?- =2.2-1.3 =0.9
b) standard deviation of this? difference =(0.32+0.42)1/2 =0.5
c)probability the small bowl contains more cereal than the large?>
d)
mean and standard deviation of the total amount of cereal in the two? bowls
mean is- 2.2+1.3=3.5
standard deviation is (0.32+0.42)1/2 =0.5
e)
f)
expected amount of cereal left in the box is =16.3-1.3-2.2=12.8
The standard deviation is (0.22+0.32+0.42)1/2 =0.54
probability = P(X>4.1) = P(Z>1.2)= 1-P(Z<1.2)= 1-0.8849= 0.1151Related Questions
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