Question

Lat x be a random variable that represents the levd of ( ucose in the blood (misgra ns per deciliter of blood after a 12 hour fast. Assume that for people under 50 years old x has distribution that is approxim ately norm al with mean μ-56 and estimated standard deviation σ-48. A test result x 40 is an indication o severe excess insulin, and medication is usually prescribed. (a) What is the probability that, on a single test, x 40? (Round your answer to four decimal places.) (b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? Hint: See Theorem 6.1 O The probability distribution of x is not normal. O The probability distribution of x is Ο The probability distribution of x is approximately normal with μί: 56 and 24.00. 0 The probability distribution of x is approximately normal with gi-56 and ơi-33.94. approximately norrnal withJG-56 and 48 that i e 40? (Round your answer to four decimal places.) t (b) for n 3 tests taken a week apart. (Round your answer to four decimal places.) d) Repeat part (b) for n 5 tests taken a week apart. (Round your answer to four decimal places) (e) Compare your answers to parts (a), (b), (c), and (d). Did the probabilities decrease as n increased? Yes No

Answer 1

$\mathbf{SOLUTION}$

$\mathbf{given:}\;mean\;(\mu)=56,\;standard\;deviation\;(\sigma)=48$ \

$\mathbf{formula:}\;z=\frac{x-\mu}{\sigma}$

$\mathbf{{\color{Blue} (a)}}\;P(x<40)\Rightarrow P\left ( \frac{x-\mu}{\sigma}<\frac{40-56}{48} \right )$

$=P(z<-0.33)$

$={\color{Red}0.3707}\;(from\;z-table)$

$\mathbf{{\color{Blue} (b)}}\;mean\;of\;the\; sampling\; distribution \;of \;\bar x,\;\mu_{\bar x}=\mu={\color{Red} 56}$

$\\standard\;deviation\;of\;the\; sampling\; distribution \;of \;\bar x,\;\sigma_{\bar x}=\frac{\sigma }{\sqrt{n}}=\frac{48}{\sqrt{2}}={\color{Red} 33.94}$

${\color{Red} The\; probability\; distribution\; of \;\bar x\; is \;approximately\; normal \;with\;\mu_{\bar x}=56 \;and\; \sigma_{\bar x}=33.94}$

$P(\bar x<40)\Rightarrow P\left ( \frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{40-56}{\frac{48}{\sqrt{2}}} \right )$

$=P(z<-0.47)$

$={\color{Red}0.3192}\;(from\;z-table)$

$\mathbf{{\color{Blue} (c)}}\;P(\bar x<40)\Rightarrow P\left ( \frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{40-56}{\frac{48}{\sqrt{3}}} \right )$

$=P(z<-0.58)$

$={\color{Red}0.2819}\;(from\;z-table)$

$\mathbf{{\color{Blue} (d)}}\;P(\bar x<40)\Rightarrow P\left ( \frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{40-56}{\frac{48}{\sqrt{5}}} \right )$

$=P(z<-0.75)$

$={\color{Red}0.2266}\;(from\;z-table)$

$\mathbf{{\color{Blue} (e)}}\;{\color{Red} Yes}$