Question

Question

A survey of 1000 people answering yes and no is taken. If the
population proportion of "yes’’ answers is *assumed* to be
0.60, what is the probability that 62% *or* more in the
survey will answer yes? (Use the normal approximation to give an
answer).

Answer #1

p = 0.60

n = 1000

P( < A) = P(Z < (A - )/)

= p = 0.60

=

=

= 0.0155

P(62% *or* more in the survey will answer yes) =
P( >
0.62)

= 1 - P( < 0.62)

= 1 - P(Z < (0.62 - 0.60)/0.0155)

= 1 - P(Z < 1.29)

= 1 - 0.9015

= **0.0985**

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