Question

The true proportion of people vaccinated against Anfluanza is known to be 62%. If a sample of 45
individuals is selected. Find the probability that the sample proportion ^p is
(a) within 0.03 of the true proportion.
(b) less than the true proportion by 0.02 or more.

Answer 1

When the sample size is sufficiently large (> 30), according to Central Limit Theorem, the distribution of sample proportion, will be approximately normal.

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

Population proportion, p = 0.62

Sample size, n = 45

= p = 0.62

=

=

= 0.0724

a) P( is within 0.03 of the true proportion) = P(0.62-0.03 < < 0.62+0.03)

= P(0.59 < < 0.65)

= P( < 0.65) - P( < 0.59)

= P(Z < 0.03/0.0724) - P(Z < -0.03/0.0724)

= P(Z < 0.414) - P(Z < -0.414)

= 0.6606 - 0.3394

= 0.3212

b) P( is less than the true proportion by 0.02 or more) = P( < 0.62-0.02)

= P( < 0.60)

= P(Z < (0.60-0.62)/0.0724)

= P(Z < -0.02/0.0724)

= P(Z < -0.276)

= 0.3913