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Sagot :
Answer:
a) P(Z > 1.09) = 0.1379
b) P(Z < -0.22) = 0.4129.
c) P(Z < -1.96) = 0.025, P(Z > -1.96) = 0.025
d) Z = 1.4
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
a. P(Z > 1.09)
This is 1 subtracted by the pvalue of Z = 1.09.
Z = 1.09 has a pvalue of 0.8621
1 - 0.8621 = 0.1379, so
P(Z > 1.09) = 0.1379
b. P(Z < -0.22)
This is the pvalue of Z = -0.22.
Z = -0.22 has a pvalue of 0.4129. So
P(Z < -0.22) = 0.4129.
c) P(Z < -1.96) or P(Z > -1.96)
Z = -1.96 has a pvalue of 0.025. So
1 - 0.025 = 0.975
P(Z < -1.96) = 0.025
P(Z > -1.96) = 0.025
d. What is the value of Z if only 8.08% of all possible Z-values are larger?
Z with a pvalue of 100 - 8.08 = 91.92 = 0.9192, so Z = 1.4
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