Please, answer these five questions with showing your work. 1. The amount of cor
ID: 2933894 • Letter: P
Question
Please, answer these five questions with showing your work.
1. The amount of corn chips dispensed into a 16-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 16.5 ounces and a standard deviation of 0.2 ounce. Suppose 100 bags of chips were randomly selected from this dispensing machine. Find the probability that the sample mean weight of these 100 bags exceeded 16.6 ounces. 2· The average score of all golfers for a particular course has a mean of 75 and a standard deviation of 4.5. Suppose 81 golfers played the course today. Find the probability that the average score of the 81 golfers exceeded 76 3. Professor Whata Guy surveyed a random sample of 420 statistics students. One of the questions was "Will you take another mathematics class?". The results showed that 252 of the students said yes. What is the sample proportion, p A of students who say they will take another math class? 4. Furnace repair bills are normally distributed with a mean of 267 dollars and a standard deviation of 20 dollars. If 64 of these repair bills are randomly selected find the probability that they have a mean cost between 267 dollars and 269 dollars. The National Association of Realtors estimates that 23% of all homes purchased in 2004 were considered investment properties. If a sample of 800 homes sold in 2004 is obtained what is the approximate probability that at most 200 homes are going to be used as investment property? 5·Explanation / Answer
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 16.5
standard Deviation ( sd )= 0.2
sample size (n) = 100
1.
P(X > 16.6) = (16.6-16.5)/0.2/ Sqrt ( 100 )
= 0.1/0.02= 5
= P ( Z >5) From Standard Normal Table
= 0
2.
mean ( u ) = 75
standard Deviation ( sd )= 4.5
sample size (n) = 81
P(X > 76) = (76-75)/4.5/ Sqrt ( 81 )
= 1/0.5= 2
= P ( Z >2) From Standard Normal Table
= 0.0228
3.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.6
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.6*0.4/420)
=0.0239
4.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 267
standard Deviation ( sd )= 20
sample size (n) = 64
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 267) = (267-267)/20/ Sqrt ( 64 )
= 0/2.5
= 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(X < 269) = (269-267)/20/ Sqrt ( 64 )
= 2/2.5 = 0.8
= P ( Z <0.8) From Standard Normal Table
= 0.7881
P(267 < X < 269) = 0.7881-0.5 = 0.2881
5.
NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.23
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.23*0.77/800)
=0.0149
P(X > 200) = (200-0.23)/0.0149
= 199.77/0.0149 = 13407.3826
= P ( Z >13407.383) From Standard Normal Table
= 0
P(X < = 200) = (1 - P(X > 200)
= 1 - 0 = 1