# Thread: Help would be appreciated..........

1. ## Help would be appreciated..........

(i)Suppose 10% of the population of a city said they agreed with a new speed limit. In a poll of 400 randomly chosen people from that city, what is the probability that fewer than 7.5% of them will say they agree with the new speed limit?

(ii)Suppose that X is a normal random variable with unknown mean and variance of 4, if Pr[X>20] = .1056, what is the mean?

2. Originally Posted by rebeccajm
(i)Suppose 10% of the population of a city said they agreed with a new speed limit. In a poll of 400 randomly chosen people from that city, what is the probability that fewer than 7.5% of them will say they agree with the new speed limit?
You could restate it as "what is the probability that <30 say they agree with the new spped limit?".

.075(400)=30

${\sigma}=\sqrt{npq}\;\ and\;\ {\mu}=np$

Use $z=\frac{x-{\mu}}{\sigma}$

Don't forget the continuity correction for x. In a binomial distribution, the possible midpoint values for "fewer than 30" are ....27, 28, 29.
Add 0.5 to the right hand boundary to get x=29.5.

Look up the z-score in the table.