contestada

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?

Respuesta :

fichoh

Answer:

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

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 Theorem

  • It states that for the sampling distribution of sample proportions of a proportion p in a sample of size n, the mean is [tex]\mu = p[/tex] and the standard deviation is [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex] .
  • The sampling distribution of sample proportions can be approximated to a normal distribution if [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].

In this problem:

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

Item a:

[tex]\mu = p = 0.005[/tex]

[tex]s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.005(0.995)}{100}} = 0.0071[/tex]

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

Item b:

[tex]np = 100(0.005) = 0.5[/tex]

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

Item c:

[tex]np = 10[/tex]

[tex]0.005n = 10[/tex]

[tex]n = \frac{10}{0.005}[/tex]

[tex]n = 2000[/tex]

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 https://brainly.com/question/25581475

Otras preguntas