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valina [46]
3 years ago
7

The lengths of songs on the radio are normally distributed with a mean length of 210 seconds, If 38.2% of all songs have lengths

between 194 and 226, then the stadard deviation of this distribution is
Mathematics
1 answer:
gladu [14]3 years ago
8 0
 i think the answer is 16 seconds                                                        
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16.A company sells calculators for $35 each. Businesses
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In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independe
enot [183]

Answer:

a)

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

b) A sample size of 991 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent

This means that \pi = 0.16

a. What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03 (round up to the next whole number).

This is n for which M = 0.03. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = z\sqrt{\frac{0.16*0.84}{n}}

0.03\sqrt{n} = z\sqrt{0.16*0.84}

\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}

(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

Question b:

99% confidence level,

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2

n = 990.2

Rounding up

A sample size of 991 is needed.

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Answer:

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