Using normal distribution to calculate $P(52.1 < bar{x} < 53.9)$
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I am given the following question in one of the lectures I was looking through.
Suppose a random sample of size $n = 400$ is to be selected from a population of size $N = 2000$. A quantitative characteristic of $x$ is of interest. Suppose that the population has mean $mu= 53$ and standard deviation $sigma= 10$. Let $bar{x}$ denote the sample average. Use the normal distribution to calculate the probability that $bar{x}$ falls between $52.1$ and $53.9$, that is, calculate $P(52.1 < bar{x} < 53.9)$
(a) $0.75$
(b) $0.87$
(c) $0.93$
(d) $0.95$
My steps were as follows:
$$Z = frac{(52.1-53)}{frac{10}{sqrt{400}}} = -1.8$$
and
$$Z = frac{(53.9-53)}{frac{10}{sqrt{400}}} = 1.8$$
I went to a $z$ score table and got the values of $.0359$ and $.9641$ respectively. I want the area between these values which is $$0.9641 - 0.0359 = 0.9282$$ so I figured the answer is (c) $0.93$, but the correct answer is (d) $0.95$.
Where am I going wrong in my calculations?
normal-distribution sampling
edited Nov 27 at 11:25
Tianlalu
3,0101938
asked Nov 27 at 7:08
shazzam
31
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up vote
0
down vote
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I am given the following question in one of the lectures I was looking through.
Suppose a random sample of size $n = 400$ is to be selected from a population of size $N = 2000$. A quantitative characteristic of $x$ is of interest. Suppose that the population has mean $mu= 53$ and standard deviation $sigma= 10$. Let $bar{x}$ denote the sample average. Use the normal distribution to calculate the probability that $bar{x}$ falls between $52.1$ and $53.9$, that is, calculate $P(52.1 < bar{x} < 53.9)$
(a) $0.75$
(b) $0.87$
(c) $0.93$
(d) $0.95$
My steps were as follows:
$$Z = frac{(52.1-53)}{frac{10}{sqrt{400}}} = -1.8$$
and
$$Z = frac{(53.9-53)}{frac{10}{sqrt{400}}} = 1.8$$
I went to a $z$ score table and got the values of $.0359$ and $.9641$ respectively. I want the area between these values which is $$0.9641 - 0.0359 = 0.9282$$ so I figured the answer is (c) $0.93$, but the correct answer is (d) $0.95$.
Where am I going wrong in my calculations?
normal-distribution sampling
edited Nov 27 at 11:25
Tianlalu
3,0101938
asked Nov 27 at 7:08
shazzam
31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given the following question in one of the lectures I was looking through.
Suppose a random sample of size $n = 400$ is to be selected from a population of size $N = 2000$. A quantitative characteristic of $x$ is of interest. Suppose that the population has mean $mu= 53$ and standard deviation $sigma= 10$. Let $bar{x}$ denote the sample average. Use the normal distribution to calculate the probability that $bar{x}$ falls between $52.1$ and $53.9$, that is, calculate $P(52.1 < bar{x} < 53.9)$
(a) $0.75$
(b) $0.87$
(c) $0.93$
(d) $0.95$
My steps were as follows:
$$Z = frac{(52.1-53)}{frac{10}{sqrt{400}}} = -1.8$$
and
$$Z = frac{(53.9-53)}{frac{10}{sqrt{400}}} = 1.8$$
I went to a $z$ score table and got the values of $.0359$ and $.9641$ respectively. I want the area between these values which is $$0.9641 - 0.0359 = 0.9282$$ so I figured the answer is (c) $0.93$, but the correct answer is (d) $0.95$.
Where am I going wrong in my calculations?
normal-distribution sampling
edited Nov 27 at 11:25
Tianlalu
3,0101938
asked Nov 27 at 7:08
shazzam
31
I am given the following question in one of the lectures I was looking through.
Suppose a random sample of size $n = 400$ is to be selected from a population of size $N = 2000$. A quantitative characteristic of $x$ is of interest. Suppose that the population has mean $mu= 53$ and standard deviation $sigma= 10$. Let $bar{x}$ denote the sample average. Use the normal distribution to calculate the probability that $bar{x}$ falls between $52.1$ and $53.9$, that is, calculate $P(52.1 < bar{x} < 53.9)$
(a) $0.75$
(b) $0.87$
(c) $0.93$
(d) $0.95$
My steps were as follows:
$$Z = frac{(52.1-53)}{frac{10}{sqrt{400}}} = -1.8$$
and
$$Z = frac{(53.9-53)}{frac{10}{sqrt{400}}} = 1.8$$
I went to a $z$ score table and got the values of $.0359$ and $.9641$ respectively. I want the area between these values which is $$0.9641 - 0.0359 = 0.9282$$ so I figured the answer is (c) $0.93$, but the correct answer is (d) $0.95$.
Where am I going wrong in my calculations?
normal-distribution sampling
normal-distribution sampling
edited Nov 27 at 11:25
Tianlalu
3,0101938
asked Nov 27 at 7:08
shazzam
31
edited Nov 27 at 11:25
Tianlalu
3,0101938
asked Nov 27 at 7:08
shazzam
31
edited Nov 27 at 11:25
Tianlalu
3,0101938
edited Nov 27 at 11:25
Tianlalu
3,0101938
edited Nov 27 at 11:25
Tianlalu
3,0101938
3,0101938
asked Nov 27 at 7:08
shazzam
31
asked Nov 27 at 7:08
shazzam
31
asked Nov 27 at 7:08
shazzam
31
31
add a comment |
add a comment |
1 Answer
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0
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You must use the finite population correction factor, because:
$$frac nN=frac{400}{2000}=0.2>0.05.$$
So, the standard deviation of the sampling distribution of the sample means (or standard error) is:
$$sigma_{bar{x}}=frac{sigma}{sqrt{n}}cdot sqrt{frac{N-n}{N-1}}$$
so you must have:
$$P(52.1 < bar{x} < 53.9)=\
Pleft(frac{52.1-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}<z<frac{53.9-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}right)approx\
P(-2.01<z<2.01)approx 0.95.$$
answered Nov 27 at 14:54
farruhota
18.7k2736
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You must use the finite population correction factor, because:
$$frac nN=frac{400}{2000}=0.2>0.05.$$
So, the standard deviation of the sampling distribution of the sample means (or standard error) is:
$$sigma_{bar{x}}=frac{sigma}{sqrt{n}}cdot sqrt{frac{N-n}{N-1}}$$
so you must have:
$$P(52.1 < bar{x} < 53.9)=\
Pleft(frac{52.1-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}<z<frac{53.9-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}right)approx\
P(-2.01<z<2.01)approx 0.95.$$
answered Nov 27 at 14:54
farruhota
18.7k2736
add a comment |
up vote
0
down vote
accepted
You must use the finite population correction factor, because:
$$frac nN=frac{400}{2000}=0.2>0.05.$$
So, the standard deviation of the sampling distribution of the sample means (or standard error) is:
$$sigma_{bar{x}}=frac{sigma}{sqrt{n}}cdot sqrt{frac{N-n}{N-1}}$$
so you must have:
$$P(52.1 < bar{x} < 53.9)=\
Pleft(frac{52.1-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}<z<frac{53.9-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}right)approx\
P(-2.01<z<2.01)approx 0.95.$$
answered Nov 27 at 14:54
farruhota
18.7k2736
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You must use the finite population correction factor, because:
$$frac nN=frac{400}{2000}=0.2>0.05.$$
So, the standard deviation of the sampling distribution of the sample means (or standard error) is:
$$sigma_{bar{x}}=frac{sigma}{sqrt{n}}cdot sqrt{frac{N-n}{N-1}}$$
so you must have:
$$P(52.1 < bar{x} < 53.9)=\
Pleft(frac{52.1-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}<z<frac{53.9-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}right)approx\
P(-2.01<z<2.01)approx 0.95.$$
answered Nov 27 at 14:54
farruhota
18.7k2736
You must use the finite population correction factor, because:
$$frac nN=frac{400}{2000}=0.2>0.05.$$
So, the standard deviation of the sampling distribution of the sample means (or standard error) is:
$$sigma_{bar{x}}=frac{sigma}{sqrt{n}}cdot sqrt{frac{N-n}{N-1}}$$
so you must have:
$$P(52.1 < bar{x} < 53.9)=\
Pleft(frac{52.1-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}<z<frac{53.9-53}{frac{10}{sqrt{400}}cdot sqrt{frac{2000-400}{2000-1}}}right)approx\
P(-2.01<z<2.01)approx 0.95.$$
answered Nov 27 at 14:54
farruhota
18.7k2736
answered Nov 27 at 14:54
farruhota
18.7k2736
answered Nov 27 at 14:54
farruhota
18.7k2736
answered Nov 27 at 14:54
farruhota
18.7k2736
18.7k2736
add a comment |
add a comment |
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