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2 years ago

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A small business ships specialty homemade candies to anywhere in the world. Past records indicate that the weight of orders is n

ormally distributed. Suppose a random sample of 16 orders is selected and each is weighed. The sample mean was found to be 110 grams with a standard deviation of 14 grams.a. Describe the sampling distribution for the sample mean.b. What is the standard error?c. For 90% confidence, what is the margin of error?d. Based on the sample results, create the 90% confidence interval and interpret.

1 answer:

FinnZ [79.3K]2 years ago

3 0

Answer with Step-by-step explanation:

Since we have given

n = 16

a) Describe the sampling distribution for the sample mean.

The sampling distribution of the mean is the population's mean from where the items are sampled.

It would be $\mu$

b. What is the standard error?

$s=\dfrac{\sigma}{\sqrt{n}}\\\\s=\dfrac{14}{\sqrt{16}}\\\\s=\dfrac{14}{4}\\\\s=3.5$

c. For 90% confidence, what is the margin of error?

Degrees of freedom = v - 1= 16-1 = 15

Using t-table, 1.753 cuts 5% level of significance in each tail.

so, Margin of error would be

$1.753\times 3.5\\\\=6.1355$

d. Based on the sample results, create the 90% confidence interval and interpret.

So, 90% confidence interval would be

$110+6.136\\\\=116.136\\\\and\\\\110-6.136\\\\=103.86$

(103.86,116.136)

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<u>Answer-</u>

<em>After 76 swings</em><em> the angle through which it swings less than 1°</em>

<u>Solution-</u>

From the question,

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So the angle of the second swing = $(30\times \frac{95}{100})^{\circ}$

Then the angle of third swing = $(30\times (\frac{95}{100})^2)^{\circ}$

So, this follows a Geometric Progression.

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So we have the nth number is the series as 1, applying the formula

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Putting the values,

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Taking logarithm of both sides,

$\Rightarrow \log \frac{1}{30} =\log (\frac{95}{100})^{n-1}$

$\Rightarrow \log \frac{1}{30} =(n-1)\log (\frac{95}{100})$

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$\Rightarrow 1.5=(n-1)(0.02)$

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