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

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SUMMARY

The calculation of the probability P(22 <= X <= 25) for a normally distributed variable X ~ N(22, 4) results in 0.4332. This is derived by standardizing the variable to a z-score using the formula z = (x - 22)/sqrt(4), which transforms the range into P(0 <= z <= 1.5). The final probability is computed as P(z <= 1.5) - P(z <= 0), yielding 0.9332 - 0.5 = 0.4332, confirming the accuracy of the result.

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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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