Binomial Distribution, 150 tosses of fair coin, P(40%<= X <=50%)

Hans Rudel

New Member
Hi, im getting the wrong answer for the following question and not sure why.

(Question) A fair coin is tossed 150 times. Find the probability that between 40% and 50% (inclusive) of the tosses will result in heads.

(I know its possible to do this proportions but i prefer to do it with approximating the binomial distribution with the normal as i find it easier to figure out the continuity correction)

np = nq = 150*0.5 = 75. Greater than 5 therefore can use normal approximation
npq = 37.5 = variance

40% of 150 = 60 tree's
50% of 150 = 75 tree's

P(60 <= X <= 75) -> continuity correction -> P(59.5 < X < 75.5)

P( ((59.5-75)/(SQRT(37.5))) < Z < ((75.5 - 75)/(SQRT(37.5))) )
= 0.9943 + 0.5032 - 1
= 0.4975

Ans should be 0.527.

I dont believe its a rounding error so if someone could point out where i have gone wrong id appreciate it.

Thanks

BGM

TS Contributor
$$\Pr\left\{Z < \frac {75.5 - 75} {\sqrt{37.5}} \right\} \approx 0.5325373$$

rather than 0.5032.

Hans Rudel

New Member
Well spotted + thank you very much for your help.

If i may ask one last question?

Question) Mr hand gained 48% of the votes in the District council elections. Find the probability that a poll of 100 randomly selected voters would show over 50% in favour of Mr Hand.

I understood this to mean the following: P(X >= 49.5) -> P(Z > ((49.5-48)/(SQRT(24.96))) )
their answer stated 0.0648 yet since the standard dev is 4.99 i have a feeling that either ive calculated the standard dev wrong or their answer is wrong.

My answer came to 0.3821 which as you can see is miles out.

Thanks very much for your time/help