Question

For a standard normal distribution, the probability of obtaining a z value between -1.9 to 1.7 is

Answer 1

P(X≤−1.9)=0.0287165598160018P(X≤−1.9)=0.0287165598160018

P(X≥1.7)=0.044565462758543P(X≥1.7)=0.044565462758543

P(−1.95996398454005≤X≤1.95996398454005)=0.95

Answer 2

Answer:

92.67% probability of obtaining a z value between -1.9 to 1.7.

Step-by-step explanation:

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

For a standard normal distribution, the probability of obtaining a z value between -1.9 to 1.7 is

This probability is the pvalue of z = 1.7 subtracted by the pvalue of z = -1.9.

z = 1.7 has a pvalue of 0.9554.

z = -1.9 has a pvalue of 0.0287.

So there is a 0.9554-0.0287 = 0.9267 = 92.67% probability of obtaining a z value between -1.9 to 1.7.