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NORMAL DISTRIBUTION

NORMAL DISTRIBUTION

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NORMAL DISTRIBUTION

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  1. NORMAL DISTRIBUTION The plant manager of a manufacturing facility is concerned about drug use among plant workers and plans to implement random drug testing. One of the tests to be applied measures the level of factor-X in blood samples. Among recent users of cocaine, the level of factor-X is normally distributed with a mean of 10.0 and a variance of 1.69. Among non-users, the level of factor-X is normally distributed with a mean of 6.75 with a variance of 2.25. The employer plans to send a warning letter to all employees with a factor-X level of x or greater, with x to be determined. Hint:You may find it helpful to sketch the 2 distributions side by side using the same X-axis. (a) Find the value of x that will ensure that 90% of recent cocaine users will be sent a warning letter.  (b) If your answer to part (a) is adopted, what proportion of non-users will also be sent warning letters?

  2. SKETCHING THE DISTRIBUTION OF NON-USER AND USER GROUPS SIDE BY SIDE PROVIDES AN IMPORTANT INSIGHT. RECENT USERS HAVE HIGHER LEVELS OF X-FACTOR IN THEIR BLOOD.

  3. Figure 1: Bottom 90% of the user group selected Figure 2: Top 90% of the user group selected We want to send warning letters to 90% of recent users.  We have 2 choices -- pick the bottom 90% (shown in Figure 1) or pick the top 90% as shown in Figure 2.  Generally students pick the bottom 90% for some reason; but this approach is incorrect because doing this will eliminate 10% users with the highest X-factor scores (users to the right of the blue line in Figure 1).  Thus, the correct choice is to pick the top 90% shown in Figure 2.

  4. Top 90% in the user group (a) Find the value of x that will ensure that 90% of recent cocaine users will be sent a warning letter. X = m + Z*s = 10-1.28*1.3 = 8.33 Sending warning letters to people with X-factor score of 8.33 or more will ensure that 90% of recent users will receive such a letter

  5. Top 90% in the user group (b) If your answer to part (a) is adopted, what proportion of non-users will also be sent warning letters? Z=(X-m)/s=(8.33-6.75)/1.5=-1.05 14.61% of non-users will be sent letters (this is the area to the right of the blue line under the non-user curve)

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