Part a)
It was given that 3% of patients gained weight as a side effect.
This means
[tex]p = 0.03[/tex]
[tex]q = 1 - 0.03 = 0.97[/tex]
The mean is
[tex] \mu = np[/tex]
[tex] \mu = 643 \times 0.03 = 19.29[/tex]
The standard deviation is
[tex] \sigma = \sqrt{npq} [/tex]
[tex] \sigma = \sqrt{643 \times 0.03 \times 0.97} [/tex]
[tex] \sigma =4.33[/tex]
We want to find the probability that exactly 24 patients will gain weight as side effect.
P(X=24)
We apply the Continuity Correction Factor(CCF)
P(24-0.5<X<24+0.5)=P(23.5<X<24.5)
We convert to z-scores.
[tex]P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 1.20) \\ = 0.051[/tex]
Part b) We want to find the probability that 24 or fewer patients will gain weight as a side effect.
P(X≤24)
We apply the continuity correction factor to get;
P(X<24+0.5)=P(X<24.5)
We convert to z-scores to get:
[tex]P(X \: < \: 24.5) = P(z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P(z \: < \: 1.20) \\ = 0.8849[/tex]
Part c)
We want to find the probability that
11 or more patients will gain weight as a side effect.
P(X≥11)
Apply correction factor to get:
P(X>11-0.5)=P(X>10.5)
We convert to z-scores:
[tex]P(X \: > \: 10.5) = P(z \: > \: \frac{10.5 - 19.29}{4.33} ) \\ = P(z \: > \: - 2.03)[/tex]
[tex] = 0.9788[/tex]
Part d)
We want to find the probability that:
between 24 and 28, inclusive, will gain weight as a side effect.
P(24≤X≤28)=
P(23.5≤X≤28.5)
Convert to z-scores:
[tex]P(23.5 \: < \: X \: < \: 28.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{28.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 2.13) \\ = 0.1494[/tex]
