Sampling Distributions

Question

A fair coin is tossed n = 80

 times. Let p̂ be the sample proportion of heads. Find P(0.41 < p̂ < 0.63).

 (Round your answer to four decimal places.)

P(0.41 < p̂ < 0.63) =

Answer 1

Hi Matthew,

Any time we flip coins, we can assume p=0.5, the proportion of heads, unless you are told the coin is weighted. Now, with sampling distributions of sample proportions (p-hat):

mu_p-hat=p, mean

sigma_p-hat=sqrt[(p(1-p))/n], standard deviation

So, here:

p=0.5

n=80, which means:

mu=0.5

sigma=sqrt(0.5*0.5)/sqrt(80)]

sigma=0.056

Now, we can use our classic equation in elementary statistics z=(x-mu)/sigma. Note that we have large n (n=80), so it is safe to assume relative normality. I will use z₁ and z₂ since we are working with two x values here. So:

z=(x₁-mu)/sigma

x₁=value we are given

mu=mean

sigma=standard deviation

Recall:

mu=0.5

sigma=0.056, calculated above

x₁=0.41, the lower bound of our range we were asked to calculate. SO:

z₁=(0.41-0.5)/0.056

z₁= -1.61

From z-table, P(Z<-1.61)=P(X<0.41)=0.0537

Keep these numbers in mind; now let's find z₂.

x₂=0.63, upper bound

mu=0.5

sigma=0.056

z₂=(0.63-0.5)/0.056

z₂=2.32

From z-table, P(Z<2.32)=P(X<0.63)=0.9898

Now, we want the probability that z falls between -1.61 and 2.32. We have the probability that z is less than both values, so we must subtract:

P(-1.61<Z<2.32)=P(0.41<x<0.63)=0.9898-0.0537=

P(0.41<X<0.63)=0.9361

I hope this helps.