Question

Side Effects for Migraine Medicine

In clinical trials and extended studies of a medication whose purpose is to reduce the pain associated with migraine headaches, 2% of the patients in the study experienced weight gain as a side effect. Suppose a random sample of 600 users of this medication is obtained.

1. Explain why you can use normal approximation to binomial distribution to approximate the probabilities below.

2. Approximate, up to 4 decimal digits, the probability that 20 or fewer users will experience weight gain as a side effect.

3. Approximate, up to 4 decimal digits, the probability that 22 or more users experience weight gain as a side effect.

4. Approximate, up to 4 decimal digits, the probability that between 20 and 30 patients, inclusive will experience weight gain as a side effect.

Answer 1

n = 600, p = 0.02

Mean, µ = n*p = 600 * 0.02 = 12

Standard deviation, σ = √(n*p*(1-p)) = √(600 * 0.02 * 0.98) = 3.4293

1. The normal distribution can be used as an approximation to the binomial distribution, under certain circumstances, namely: If X ~ B(n, p) and if n is large.

As both the conditions are satisfied we will use normal approximation to binomial distribution.

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2. Probability that 20 or fewer users will experience weight gain as a side effect, P(X ≤ 20) =

Using continuity correction :

P(X ≤ 20+0.5)

= P((X - µ)/σ ≤ (20.5 - 12)/3.4293)

= P(z ≤ 2.4787)

Using excel function:

= NORM.S.DIST(2.4787, 1)

= 0.9934

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3. Probability that 22 or more users experience weight gain as a side effect, P(X ≥ 22) =

Using continuity correction :

P(X ≥ 22-0.5)

= P((X - µ)/σ ≥ (21.5 - 12)/3.4293)

= P(z ≥ 2.7703)

= 1 - P(z < 2.7703)

Using excel function:

= 1 - NORM.S.DIST(2.7703, 1)

= 0.0028

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4. Probability that between 20 and 30 patients, inclusive will experience weight gain as a side effect=

P(20 ≤ X ≤ 30)

Using continuity correction :

P(20-0.5 ≤ X ≤ 30+0.5)

= P((19.5 - 12)/3.4293 ≤ (X - µ)/σ ≤ (30.5 - 12)/3.4293)

= P( 5.3947 ≤ z ≤ 2.187)

= P(z < 5.3947) - P(z < 2.187)

Using excel function:

= NORM.S.DIST(5.3947, 1) - NORM.S.DIST(2.187, 1)

= 0.0144