Question

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.

Answer 1

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

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

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

$SD = \sqrt{72.1344}$

$SD = 8.49319727782$

$SD = 8.4932$ --- approximated

Hence, the calculated standard deviation is approximately 8.4932