Question 67265: 1. The average yearly cost per household of owning a dog is $186.80. Suppose that we randomly select 50 households that own a dog. What is the probability that the sample mean for these 50 households is less than $175.00.
Assume o = $32
z =X - u/o
z = 175.00- 186.80/32 = -11.8/32 = -3.6875
0.5000-0.1406 = 0.6406
P(X < 175) = 0.6406 = 64% (not sure if used correct steps)
b. Sample Mean - mean/ standard deviation/ square root of N.
175 - 186.80 / 32/(squqare root of 50 = -2.607
0.5000 +4955 = .9955 = 99% or 100%
2. The average teacher’s salary in North Dakota is $29863. Assume a normal distribution with = $5100.
a. What is the probability that a randomly selected teacher’s salary is greater than $40,000?
b. What is the probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000
a.
z = 29863 - 40000/5100 = -10137/5100 = -1.98 0.5000+.4761 = .9761 = 98%
P(X>40,000)
3. For a certain year the average annual salary in Pennsylvania was $24,393. Assume that salaries were normally distributed for a certain group of wage earners, and the standard deviation of this group was $4362.
a. Find the probability that randomly selected individual earned less than $26,000.
b. Find the probability that, for a randomly selected sample of 25 individuals, the mean salary was less than $26,000.
c. Why is the probability for Part b higher than the probability for Part a?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. The average yearly cost per household of owning a dog is $186.80. Suppose that we randomly select 50 households that own a dog. What is the probability that the sample mean for these 50 households is less than $175.00.
Assume o = $32
z =X - u/o
COMMENT: If sigma=$32 , s=32/sqrt50 (this is the std dev of the sample means)
z = [175.00- 186.80]/[32sqrt50] = -11.8/4.525 = -2.607
P(z<-2.607)=0.004567
P(X < 175) = 0.4%
b. Sample Mean - mean/ standard deviation/ square root of N.
COMMENT: You have it right below.
175 - 186.80 / 32/(square root of 50 = -2.607
But you want P(z<-2.607) which is approx 0.4%
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2. The average teacher’s salary in North Dakota is $29863. Assume a normal distribution with sigma = $5100.
a. What is the probability that a randomly selected teacher’s salary is greater than $40,000?
b. What is the probability that the mean for a sample of 80 teachers’ salaries is greater than $30,000
a.
COMMENT: You want the z-value of 40,000. You need to switch your numbers.
z(40,000) = (40000-29863)/5100 = 10137/5100 = 1.98
P(X>40,000)=P(z>1.98)=0.0239 or appox 2.4%
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3. For a certain year the average annual salary in Pennsylvania was $24,393. Assume that salaries were normally distributed for a certain group of wage earners, and the standard deviation of this group was $4362.
a. Find the probability that randomly selected individual earned less than $26,000.
COMMENT: Find the z of 26000
z(26000)=[26000-24393]/4362=0.3684
P(z<0.3684)=0.6437... or appox 64.4%
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b. Find the probability that, for a randomly selected sample of 25 individuals, the mean salary was less than $26,000.
COMMENT: Here you have to change the std dev to 24393/sqrt(25)
c. Why is the probability for Part b higher than the probability for Part a?
COMMENT: The standard deviation is smaller resulting in a higher z-value
and therefore more population or probability below that z value.
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Cheers,
Stan
PS: If you have more stat questions as you go through your course I would
be happy to help you. I am looking for an opportunity to review what I
know about stat. You may email me directly at stanbon@comcast.net
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