Find the following z values for the standard normal variable Z. Use Table 1. (Negative values should be indicated by a minus sign. Round your answers to 2 decimal places.) |
a. | P(Z ≤ z) = 0.9478 | |
b. | P(Z > z) = 0.6140 | |
c. | P(−z ≤ Z ≤ z) = 0.80 | |
d. | P(0 ≤ Z ≤ z) = 0.2392 | |
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S.No. | Required Probability | Required Z-Score |
a. | P(Z ≤ z) = 0.9478 | 1.623 |
b. | P(Z > z) = 0.6140 | -0.292 |
c. | P(−z ≤ Z ≤ z) = 0.80 | 1.281 |
d. | P(0 ≤ Z ≤ z) = 0.2392 | 0.638 |
I calculated these values using the combination of goal seek function of MS-Excel and the function "=NORM.S.DIST(p-value,TRUE)" by aksing the goal seek function to calculate the required probabilities.
(a)
(b)
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I hope this solves your doubt.
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