1) It is known 30% of university students get flu shots. Consider the 130 studen

Question

1) It is known 30% of university students get flu shots. Consider the 130 students in STAT class a simple random sample (SRS) of university students. What is the probability that more than half of university students have or will get a flu shot this season? What assumptions do you need to compute this probability?

2) Suppose that 35% of Canadian households have a dog. Your street has 50 houses. Compute the probability that less than 10 households on your street have a dog? What assumptions do you need to compute this probability?

Explanation / Answer

Question 1

We are given

n = 130, p = 30% = 0.30, q = 1 – p = 1 – 0.30 = 0.70

We have to check assumptions n*p > 5 and n*q > 5 for using normal approximation to binomial distribution.

n*p = 130*0.30 = 39 > 5

n*q = 130*0.70 = 91 > 5

Both assumptions satisfied. We can use normal approximation to binomial distribution.

We have to find P(X>half students) = P(X>130/2) = P(X>65)

We have to compute P(X>65)

By adding continuity correction 0.5, we have to find P(X>65.5)

P(X>65.5) = 1 – P(X<65.5)

Z = (X – Mean) / SD

Mean = n*p = 130*0.30 = 39

SD = sqrt(n*p*q) = sqrt(130*0.30*0.70) = 5.22494

Z = (65.5 – 39) / 5.22494

Z = 5.071828

P(Z< 5.071828) = P(X<65.5) = 1

P(X>65.5) = 1 – P(X<65.5)

P(X>65.5) = 1 – 1 = 0.00

Required probability = 0.00

Question 2

We are given

n = 50

p = 0.35

q = 1 – 0.35 = 0.65

We have to find P(X<10)

n*p = 50*0.35 = 17.5 > 5

n*q = 50*0.65 = 32.5 > 5

Both assumptions satisfied. We can use normal approximation to binomial distribution.

Mean = n*p = 50*0.35 = 17.5

SD = 3.372684

Z = (X – Mean) / SD

Z = (10 – 17.5) / 3.372684

Z = -2.22375

P(Z< -2.22375) = P(X<10) = 0.013083

Required probability = 0.013083

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