4. + Question Details My Notes Ask Your Teacher Penny uses powdered peanut butte

Question

4. + Question Details My Notes Ask Your Teacher Penny uses powdered peanut butter(PB). On the PB jar it says that the serving size is two tablespoons which is 12 grams. Measuring out a tablespoon is hard so the actual amount of PB in a measured serving(actual serving) has an expected value of 12.3 grams and a standard deviation 1.2 grams of PB. It is known that PB serving size has a normal distribution. Penny takes one actual serving of PB. Let X = the number of grams of PB in the actual serving. a) Find the probability that a random actual serving of PB weighs less than 12 grams b) Calculate the variance of X c) Find the probability that X is within 0.5 standard deviations of its expected value? d) What is the 12th percentile of X? e) Suppose Penny takes an actual serving of PB on each day of the week (7 days). Let S be the total number of grams of PB in Pennys servings. If Penny needs at least 84 grams of PB to meet her weekly health goal then what is the probability that Penny meets her goal? Assume independence between servings. t)What is the standard deviation of the total S from part e) 9) Suppose Penny takes an actual serving of PB on each day of the week (7 days). What is the probability that every one of the servings has at least 12 grams of PB? h) Add any comment to the text box below.

Explanation / Answer

X: Number of grams of PB in actual serving.

It is given that E(X) = 12.3 and sd(X) = 1.2.

a) P(X < 12) = P((X - 12.3)/1.2 < (12 - 12.3)/1.2) = P(Z < -0.25) = Phi(-0.25) = 0.4013.

b) Variance(X) = (1.2)^2 = 1.44.

c) P(12.3 - 0.25*sd < X < 12.3 + 0.25*sd) = P(-0.25 < Z < 0.25) = Phi(0.25) - Phi(-0.25) = 0.1974.

d) If the 12th percentile of X be p then,

P(X < p) = 0.12, i.e. Phi((p - 12.3)/1.2) = 0.12 , i.e. (p - 12.3)/1.2 = -1.175.

p = -1.175*1.2 + 12.3 = 10.89.

e) Here Y = 7X. So, Y follows a normal distribution with mean E(Y) = 7*12.3 = 86.1 and standard deviations sy=7*( 1.2^2) = 10.08.

Now, the probability of (Y > 84) = P(Z > (84-86.1)/sqrt(sy)) = 1 - Phi(0.6614) = 0.2523.

f) Standard deviation of the total S = sqrt(10.08) = 3.1749.

g) If each days are independent, then each days serving at least 12 gms = {P(X>12)}^7 = (1-0.4013)^7 = 0.0276.

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