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3 years ago

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Consider the following binomial experiment: A newspaper publisher claims that 52% of the people in a certain community read his

newspaper. Doubting the assertion, a competitor randomly surveys 289 people in the community. Find the standard deviation of this experiment, if the newspaper publisher's claim is correct.
a) 8.1932
b) 8.3932
c) 8.4932
d) 8.6932
e) 8.5932
f) None of the above.

1 answer:

AURORKA [14]3 years ago

4 0

Answer:

$SD = 8.4932$

Step-by-step explanation:

Given

$Sample\ Size (n) = 289$

$proportion, p= 52\%$

Required

Determine the standard deviation (SD)

The standard deviation of a binomial distribution is calculated as thus:

$SD = \sqrt{np(1-p)}$

Substitute values for n and p

$SD = \sqrt{289 * 52\% * (1-52\%)}$

Convert to decimals

$SD = \sqrt{289 * 0.52 * (1-0.52)}$

$SD = \sqrt{289 * 0.52 * 0.48}$

$SD = \sqrt{72.1344}$

$SD = 8.49319727782$

$SD = 8.4932$ --- <em>approximated</em>

<em>Hence, the calculated standard deviation is approximately 8.4932</em>

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Read 2 more answers

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The semiperimeter of the triangle is:

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Hope this helps!

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3 years ago