MHB Calculate using Standard Normal Table: P(22 <= X <= 25) = .4332

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The discussion focuses on calculating the probability P(22 <= X <= 25) using the standard normal table for a normal distribution X ~ N(22,4). The transformation to the standard normal variable z is performed, resulting in z = (x - 22)/2. The probability is calculated as P(0 <= z <= 1.5), which is derived from the standard normal table values: P(z <= 1.5) = 0.9332 and P(z <= 0) = 0.5. The final result is P(22 <= X <= 25) = 0.4332, confirming the calculations are correct. The thread emphasizes the importance of using the standard normal table accurately for such probability calculations.
Uniman
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Using the standard normal table in your lecture notes, calculate to four decimal places

Table

file:///Users/manukumarindia/Desktop/Screen%20Shot%202012-10-29%20at%2011.12.15%20AM.png

Work done so far.

If X ~ N(22,4), z = (x - 22)/sqrt(4) ~ N(0,1) Thus, P(22 <= X <= 25) = P((22-22)/2 <= z <= (25-22)/2) = P(0 <= z <= 1.5) = P(z <= 1.5) - P(Z <= 0) Now, we use the table above. P(z <= 1.5) - P(Z <= 0) = .9332 - .5 = .4332
 
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Uniman said:
Using the standard normal table in your lecture notes, calculate to four decimal places

Table

file:///Users/manukumarindia/Desktop/Screen%20Shot%202012-10-29%20at%2011.12.15%20AM.png

Work done so far.

If X ~ N(22,4), z = (x - 22)/sqrt(4) ~ N(0,1) Thus, P(22 <= X <= 25) = P((22-22)/2 <= z <= (25-22)/2) = P(0 <= z <= 1.5) = P(z <= 1.5) - P(Z <= 0) Now, we use the table above. P(z <= 1.5) - P(Z <= 0) = .9332 - .5 = .4332



That seems OK.

CB
 
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