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?

Using the Central Limit Theorem, it i found that:

In this problem:

Item a:

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

Item b:

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

Item c:

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

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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

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step-by-step explanation:

57.09 minus 7.0435 equals

step-by-step explanation:

just subtrach them both and u get the answer of 50.0465