Question

Given a standardized normal distribution (with a mean of O and a standard deviation of 1), determine the following probabilities.
a. P(Z > 1.09) b. P(Z < -0.22) c. P(-1.96 d. What is the value of Z if only 8.08% of all possible Z-values are larger?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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