Let x be a random variable that represents the level of gluc

Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean = 50 and estimated standard deviation = 35. A test result x < 40 is an indication of 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.

What is the probability that x < 40? (Round your answer to four decimal places.)

(c) Repeat part (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.)

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    40      
u = mean =    50      
          
s = standard deviation =    35      
          
Thus,          
          
z = (x - u) / s =    -0.285714286      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.285714286   ) =    0.387548481 [ANSWER]

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b)

It will have a new standard deviation, which is 1/sqrt(2) of the original.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    40      
u = mean =    50      
n = sample size =    2      
s = standard deviation =    35      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -0.404061018      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.404061018   ) =    0.343083925 [ANSWER]

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c)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    40      
u = mean =    50      
n = sample size =    3      
s = standard deviation =    35      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -0.494871659      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.494871659   ) =    0.310345359 [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    40      
u = mean =    50      
n = sample size =    5      
s = standard deviation =    35      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -0.638876565      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -0.638876565   ) =    0.261451617 [ANSWER]