A pizza company advertises that it puts 0.5 pounds of real mozzarella cheese on

Question

A pizza company advertises that it puts 0.5 pounds of real mozzarella cheese on its medium pizzas. Suppose that the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pounds and a standard deviation of 0.005 pounds. (Let y = the amount of cheese on a medium pizza. Round the answers to three decimal places.)

(a) What is the probability that the amount of cheese on a medium pizza is between 0.505 and 0.510 pounds? P(0.505 y 0.510) = Incorrect: Your answer is incorrect.

(b) What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? P(y > + 2) = Correct: Your answer is correct.

(c) What is the probability that two randomly selected medium pizzas all have at least 0.495 pounds of cheese? Correct: Your answer is correct.

Explanation / Answer

Answer:

A pizza company advertises that it puts 0.5 pounds of real mozzarella cheese on its medium pizzas. Suppose that the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pounds and a standard deviation of 0.005 pounds. (Let y = the amount of cheese on a medium pizza. Round the answers to three decimal places.)

(a) What is the probability that the amount of cheese on a medium pizza is between 0.505 and 0.510 pounds? P(0.505 y 0.510) = Incorrect: Your answer is incorrect.

Z value for 0.505, z =(0.505-0.5)/0.005 =1

Z value for 0.510, z =(0.510-0.5)/0.005 =2

P(0.505 y 0.510) = P( 1<z<2) P( z <2) –P( z <1)

= 0.9772 - 0.8413

=0.1359

=0.136 ( three decimals)

(b) What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? P(y > + 2) = Correct: Your answer is correct.

P(y > + 2) = P( y>0.51)

Z value for 0.51 is 2

P( z>2 = 0.0228

=0.023

(c) What is the probability that two randomly selected medium pizzas all have at least 0.495 pounds of cheese? Correct: Your answer is correct.

Z value for 0.495, z =(0.495-0.5)/0.005 =-1

P(x >0.495) = P( z > -1)= 0.8413

Both have at least 0.0495 pounds = 0.8413*0.8413 =0.70778

=0.708

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