Hi Anni,
Use the normal approximation to the binomial distribution or technology (TI-83 Plus or statistical software) For the normal approximation, we first need the mean and standard deviation:
mu=np
sigma= sqrt(npq)
n=total sample size
p=probability of success--in this case, dog is spayed or neutered
q=probability of failure = 1-p
mu= 39*0.7 = 27.3
sigma= sqrt(39*0.7*0.3) = 2.862
a. Continuity correction is required for normal approximation and this requires us to adjust the x-value by 0.5 either way, so we want:
P(28.5<x<29.5)
We can use z scores for this.
z1 = (x1-mu)/sigma
x1 = 28.5
mu= 27.3
sigma = 2.862
z1 = (28.5-27.3)/2.862 = 0.42
z2 = (x2-mu)/sigma
x2 = 29.5
mu= 27.3
sigma = 2.862
z1 = (29.5-27.3)/2.862 = 0.77
P(0.42<z<0.77)=P(28.5<x<29.5)
P(z<0.42) = 0.6628
P(z<0.77) = 0.7794
We want the range between so subtract:
P = 0.7794 - 0.6628
P = 0.1166
b. This question asks for at most, so we will approximate by adding 0.5 only. We want:
P(X<=27.5) We can use our z equation again:
z= (x-mu)/sigma
x=27.5
mu=27.3
sigma=2.862
z= (27.5-27.3)/2.862
z= 0.07
P(Z<0.07)=0.5279
P = 0.5279
c. This time, it's at least, so we correct down. We want:
P(X>=27.5)
We computed the z-score for x=27.5 above. It was:
z=0.07
P(z<0.07) = 0.5279
But remember we want at least, which implies greater than or equal to, so we apply the Complement Rule aka the "1 minus trick:"
P(z>0.07)=P(X>27.5)=1-0.5279
P = 0.4721
d. This time we have to subtract on the low end and add on the high end. We want:
P(24.5<x<31.5)
We can use the classic z=(x-mu)/sigma for both of these.
x1 = 24.5
mu=27.3
sigma=2.862
z1 = (24.5 - 27.3)/2.862 = -0.98
x2 = 31.5
mu=27.3
sigma=2.862
z2 = (31.5 - 27.3)/2.862 = 1.47
P(z< -0.98) = 0.1635
P(z<1.47) = 0.9292
We want the area between, so we subtract:
P = 0.9292 - 0.1635
P = 0.7657
I hope this helps.
