Problem

The Orange County Register, as part of its Sunday health supplement, reporfed that 64 perc...

The Orange County Register, as part of its Sunday health supplement, reporfed that 64 percent of American men over the age of 18 consider nutrition a top priority in their lives. Suppose we select a sample of 60 men. What is the likelihood that:

a.   32 or more consider nutrition important?


b.   44 or more consider nutrition important?


c.   More than 32 but fewer than 43 consider nutrition important?


d.   Exactly 44 consider diet important?

Step-by-Step Solution

Solution 1

Mean

$$ \begin{aligned} \mu &=n \pi \\ &=60 \times 0.64 \\ \mu &=38.4 \end{aligned} $$

Standard deviation of the random variable

$$ \begin{aligned} \sigma &=\sqrt{n \pi(1-\pi)} \\ &=\sqrt{60 \times 0.64 \times(1-0.64)} \\ &=\sqrt{60 \times 0.64 \times 0.36} \\ &=\sqrt{13.824} \\ \sigma &=3.72 \end{aligned} $$

a)$$ \begin{aligned} &=P(x \geq 31.5) \\ &=P\left(\frac{x-\mu}{\sigma} \geq \frac{31.5-38.4}{3.72}\right) \\ &=P(z \geq-1.85) \\ &=0.5+P(z<1.85) \\ &=0.5+0.4678 \\ &=0.9678 \end{aligned} $$

b)

$$ \begin{aligned} &=P(x \geq 43.5) \\ &=P\left(\frac{x-\mu}{\sigma} \geq \frac{43.5-38.4}{3.72}\right) \\ &=P(z \geq 1.37) \\ &=0.5-P(z<1.37) \\ &=0.5-0.4147 \\ &=0.0853 \end{aligned} $$

c) $$ \begin{aligned} &=P(32.5 \leq x \leq 42.5) \\ &=P\left(\frac{32.5-38.4}{3.72} \leq \frac{x-\mu}{\sigma} \leq \frac{42.5-38.4}{3.72}\right) \\ &=P(-1.58 \leq z \leq 1.10) \\ &=P(z \leq 1.58)+P(z \leq 1.10) \\ &=0.4429+0.3643 \\ &=0.8072 \end{aligned} $$

d)

$$ \begin{aligned} &=P(x>44.5)-P(x>43.5) \\ &=P\left(\frac{x-\mu}{\sigma} \geq \frac{44.5-38.4}{3.72}\right)-P\left(\frac{x-\mu}{\sigma} \geq \frac{43.5-38.4}{3.72}\right) \\ &=P(z \geq 1.64)-P(z \geq 1.37) \\ &=0.4495-0.4147 \\ &=0.0348 \end{aligned} $$

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