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3 years ago

I neeed this rithg now

Mathematics

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LUCKY_DIMON [66]3 years ago

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Answer:

Step-by-step explanation:

The farther left you get from zero on the number line the less the number is.

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In studies for a medication, 3 percent of patients gained weight as a side effect. Suppose 643 patients are randomly selected.

timofeeve [1]

Part a)

It was given that 3% of patients gained weight as a side effect.

This means

$p = 0.03$

$q = 1 - 0.03 = 0.97$

The mean is

$\mu = np$

$\mu = 643 \times 0.03 = 19.29$

The standard deviation is

$\sigma = \sqrt{npq}$

$\sigma = \sqrt{643 \times 0.03 \times 0.97}$

$\sigma =4.33$

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.

$P(23.5 \: < \: X \: < \: 24.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 1.20) \\ = 0.051$

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:

$P(X \: < \: 24.5) = P(z \: < \: \frac{24.5 - 19.29}{4.33} ) \\ = P(z \: < \: 1.20) \\ = 0.8849$

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:

$P(X \: > \: 10.5) = P(z \: > \: \frac{10.5 - 19.29}{4.33} ) \\ = P(z \: > \: - 2.03)$

$= 0.9788$

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:

$P(23.5 \: < \: X \: < \: 28.5) = P( \frac{23.5 - 19.29}{4.33} \: < \: z \: < \: \frac{28.5 - 19.29}{4.33} ) \\ = P( 0.97\: < \: z \: < \: 2.13) \\ = 0.1494$