Question

1. Costs for standard veterinary services at a local animal hospital follow a Normal distribution with a mean of $89 and a standard deviation of $23. What is the probability that one bill for veterinary services costs between $45 and $132?

	a.	0.5000
	b.	0.0586
	c.	0.5293
	d.	0.4707
	e.	0.9414

2.

Lloyd's Cereal company packages cereal in 1 pound boxes (16 ounces). A sample of 49 boxes is selected at random from the production line every hour, and if the average weight is less than 15 ounces, the machine is adjusted to increase the amount of cereal dispensed. If the mean for 1 hour is 1 pound and the standard deviation is 0.2 pound, what is the probability that the amount dispensed per box will have to be increased?

	a.	0.2144
	b.	0.0287
	c.	0.9856
	d.	0.3773
	e.	0.0144

Answer 1

Part 1)

X ~ N ( µ = 89 , σ = 23 )
P ( 45 < X < 132 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 45 - 89 ) / 23
Z = -1.913
Z = ( 132 - 89 ) / 23
Z = 1.8696
P ( -1.91 < Z < 1.87 )
P ( 45 < X < 132 ) = P ( Z < 1.87 ) - P ( Z < -1.91 )
P ( 45 < X < 132 ) = 0.9692 - 0.0279
P ( 45 < X < 132 ) = 0.9414

Part 2)

X ~ N ( µ = 16 , σ = 3.2 )
P ( X̅ < 15 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 15 - 16 ) / ( 3.2 / √49 )
Z = -2.1875
P ( ( X - µ ) / ( σ/√(n)) < ( 15 - 16 ) / ( 3.2 / √(49) )
P ( X̅ < 15 ) = P ( Z < -2.19 )
P ( X̅ < 15 ) = 0.0144