In: Statistics and Probability
A physical education professor claims that 35% of third-grade children can do a handstand. If this claim is true:
a. Find the probability that 10 or more third-grade children out of a random sample of 25 can do a handstand.
i. Use the exact binomial distribution.
ii. Use the normal distribution without a continuity correction.
iii. Use the normal distribution with a continuity correction.
b. Find the probability that 40 or more third-grade children out of a random sample of 100 can do a handstand.
i. Use the normal distribution without a continuity correction.
ii. Use the normal distribution with a continuity correction.
c. Based on the results of parts a and b, is the correction for continuity more important in large or in small samples?
Part a.i.
(by using exact binomial distribution)
We are given
n = 25, p = 0.35
We have to find P(X≥10)
P(X≥10) = 1 – P(X<10)
P(X≥10) = 1 – P(X≤9)
P(X≥10) = 1 – 0.630309
(by using excel or binomial table)
P(X≥10) = 0.369691
Required probability = 0.369691
Part a.ii.
(by using normal approximation without continuity correction)
We are given
n = 25, p = 0.35, q = 1 – p = 1 – 0.35 = 0.65
Mean = np = 25*0.35 = 8.75
SD = sqrt(npq) = sqrt(25*0.35*0.65) = 2.384848
We have to find P(X≥10)
P(X≥10) = 1 – P(X<10)
Z = (X – mean)/Sd
Z = (10 - 8.75)/ 2.384848
Z = 0.524142
P(Z<0.524142) = P(X<10) = 0.69991
(by using z-table)
P(X≥10) = 1 – P(X<10)
P(X≥10) = 1 – 0.69991
P(X≥10) = 0.30009
Required probability = 0.30009
Part a.iii.
(by using normal approximation with continuity correction)
We are given
n = 25, p = 0.35, q = 1 – p = 1 – 0.35 = 0.65
Mean = np = 25*0.35 = 8.75
SD = sqrt(npq) = sqrt(25*0.35*0.65) = 2.384848
We have to find P(X≥10) = P(X>9.5) (by using continuity correction)
P(X>9.5) = 1 – P(X<9.5)
Z = (X – mean)/Sd
Z = (9.5 - 8.75)/ 2.384848
Z = 0.314485
P(Z<0.314485) = P(X<9.5) = 0.623424
(by using z-table)
P(X≥10) = 1 – 0.623424
P(X≥10) = 0.376576
Required probability = 0.376576
Part b.i.
(by using normal approximation without continuity correction)
We are given
n = 100, p = 0.35
We have to find P(X≥40)
We are given
n = 100, p = 0.35, q = 1 – p = 1 – 0.35 = 0.65
Mean = np = 100*0.35 = 35
SD = sqrt(npq) = sqrt(100*0.35*0.65) = 4.769696
We have to find P(X≥40)
P(X≥40) = 1 – P(X<40)
Z = (X – mean)/Sd
Z = (40 - 35)/ 4.769696
Z = 1.048285
P(Z<1.048285) = 0.852746
(by using z-table)
P(X≥40) = 1 – P(X<40)
P(X≥40) = 1 – 0.852746
P(X≥40) = 0.147254
Required probability = 0.147254
Part b.ii.
(by using normal approximation with continuity correction)
We are given
n = 100, p = 0.35
We have to find P(X≥40) = P(X>39.5)
We are given
n = 100, p = 0.35, q = 1 – p = 1 – 0.35 = 0.65
Mean = np = 100*0.35 = 35
SD = sqrt(npq) = sqrt(100*0.35*0.65) = 4.769696
P(X>39.5) = 1 – P(X<39.5)
Z = (X – mean)/Sd
Z = (39.5 - 35)/ 4.769696
Z = 0.943456
P(Z<0.943456) = 0.827276
(by using z-table)
P(X≥40) = 1 – 0.827276
P(X≥40) = 0.172724
Required probability = 0.172724
Part c
The correction for continuity is more important in small samples, because there is more difference in probabilities for using continuity and without using continuity correction for small samples.