Question

In: Statistics and Probability

A physical education professor claims that 35% of third-grade children can do a handstand. If this...

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?

Solutions

Expert Solution

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.


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