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Anna71 [15]
3 years ago
10

22a) PLEASE HELP BEST ANSWER BRAINLIEST

Mathematics
1 answer:
adoni [48]3 years ago
8 0

Answer:

the answer is k=1

Step-by-step explanation:

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Suppose we have a population whose proportion of items with the desired attribute is p = 0:5. (a) If a sample of size 200 is tak
crimeas [40]

Answer:

a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

Step-by-step explanation:

This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.

The parameters for the normal distribution will be

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/200}= 0.0353

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.0353}=-0.85\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.0353}=0.28

We can now calculate the probabilities:

P(0.47

If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.

b) If the sample size change, the standard deviation of the normal distribution changes:

\mu=p=0.5\\\\\sigma=\sqrt{p(1-p)/n} =\sqrt{0.5*0.5/100}= 0.05

We can calculate the z values for x1=0.47 and x2=0.51:

z_1=\frac{x_1-\mu}{\sigma}=\frac{0.47-0.5}{0.05}=-0.6\\\\z_2=\frac{x_2-\mu}{\sigma}=\frac{0.51-0.5}{0.05}=0.2

We can now calculate the probabilities:

P(0.47

If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.

8 0
3 years ago
How can 38/6 be expressed as a decimal?
seraphim [82]
38/6 = 19/3
divide 19 by 3 :-

= 6.333....

The 3 is recurring
8 0
3 years ago
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