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3 years ago

BAKERY Rolls are being prepared to go to grocery stores, Divide 72 rolls

Mathematics

1 answer:

DedPeter [7]3 years ago

3 0

Answer:

Step-by-step explanation:

72 divided by 2 = 36

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lesantik [10]

Answer:

a) The sampling distribution for n=540 has a mean sample proportion of p=0.37 and a standard deviation of σs=0.0208.

b) probability = 0.99998

d) probability = 0.99874

e) You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

Step-by-step explanation:

a) We have a known population proportion π=0.37 and we have to describe the sampling distribution when the sample size is n=540.

The mean sample proportion is expected to be the same as the population proportion:

$\bar p = \pi = 0.37$

The standard deviation of the sampling will be the population standard deviation divided by the square root of the sample size:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{540}}=\sqrt{0.000431667}=0.0208$

Then, we can say that the sampling distribution will have a p=0.37 and a standard deviation σs=0.0208.

b) We have to calculate the probability that the sample proportion will be within 0.09 of the population proportion.

We can calculate the z-value as:

$z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0208}=4.3269\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0208}=-4.3269$

As the distribution is symmetrical, we can calculate the probabilty that he sample proportion will be within 0.09 of the population proportion as:

$P(|p-\bar p|$

probability = 0.99998

d. Now the sample is smaller (n=300), so the standard deviation of the samping distribution:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{300}}=\sqrt{0.000777}=0.0279$

We have to recalculate the z-scores:

$z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0279}=3.2258\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0279}=-3.2258$

And the probability is:

$P(|p-\bar p|$

probability = 0.99874

e. The increase in sample size is 80%

$\Delta n\%=\dfrac{n_1}{n_2}-1=\dfrac{540}{300}-1=1.8-1=0.8=80\%$

and the increase in probability is 0.12%

$\Delta P\%=\dfrac{P_1}{P_2}-1=\dfrac{0.99998}{0.99874}-1=1.0012-1=0.0012=0.12\%$

You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

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