Mathematics, 25.01.2021 20:20 korar8737

A certain chromosome defect occurs in only 1 in 200 adult Caucasian males. A random sample of 100 adult Caucasian males will be selected. The proportion of men in this sample who have the defect, ^p, will be computed. A. What is the mean value of the sample proportion ^p, and what is the standard deviation of the sample proportion?
B. Does ^p have approximately a normal distribution in this case? Explain.
A. Yes, because np < 10 and n(1 − p) < 10
B. Yes, because np > 10 and n(1 - p) > 10.
C. No, because np < 10.
D. No, because np > 10.
C. What is the smallest value of n for which the sampling distribution of ^p is approximately normal?

Answers: 2

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Answer from: karlyisaunicorn

Using the Central Limit Theorem, it i found that:

A. The mean is of 0.005, with a standard deviation of 0.0071.B. C. No, because np < 10.C. The smallest value of n for the sampling distribution to be approximately normal is 2000.Central Limit TheoremIt states that for the sampling distribution of sample proportions of a proportion p in a sample of size n, the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$ .The sampling distribution of sample proportions can be approximated to a normal distribution if $np \geq 10$ and $n(1 - p) \geq 10$ .

In this problem:

The defect occurs in only 1 in 200 adult Caucasian males, hence $p = \frac{1}{200} = 0.005$ .A sample of 100 is selected, hence n = 100

Item a:

$\mu = p = 0.005$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.005(0.995)}{100}} = 0.0071$

The mean is of 0.005, with a standard deviation of 0.0071.

Item b:

np = 100(0.005) = 0.5

np < 10, hence not normal, and option C is correct.

Item c:

np = 10

0.005n = 10

$n = \frac{10}{0.005}$

n = 2000

The smallest value of n for the sampling distribution to be approximately normal is 2000.

To learn more about the Central Limit Theorem, you can take a look at link

Answer from: OinkFred

0.005 `; 0.00499 ;

No, because np < 10 ;

2000

Step-by-step explanation:

Given that:

Number of samples , n = 100

Proportion, p = x / n

p = 1 / 200

= 0.005

p = μ

Standard deviation of sample proportion :

σp = sqrt((p(1 - p)) / n)

σp = sqrt((0.005(1 - 0.005)) / 200)

σp = sqrt((0.005(0.995)) / 200)

σp = sqrt(0.004975 / 200)

σp = sqrt(0.000024875)

σp = 0.0049874

σp = 0.00499

np = 100 * 0.005 = 0.5

n(1 - p) = 100(1-0.05) = 95

Smallest value of n for which sampling distribution is approximately normal

np ≥ 10

0.005n ≥ 10

To obtain the smallest value of n,

0.005n = 10

n = 10 / 0.005

n = 2000

Answer from: Quest

d i think

step-by-step explanation:

Me plz i just need this one last test to pass

Answer from: Quest

proofs involving isosceles triangles often require special consideration because an isosceles triangle has several distinct properties that do not apply to normal triangles.(more about triangle types) therefore, when you are trying to prove that two triangles are congruent, and one or both triangles, are isosceles you have a few theorems that you can use to make your life easier.

step-by-step explanation:

an isosceles triangle has two congruent sides and two congruent angles. the congruent angles are called the base angles and the other angle is known as the vertex angle. ∠bac and ∠bca are the base angles of the triangle picture on the left. the vertex angle is ∠abc