Respuesta :
Answer:
Z-scores between -0.86 and 0.86 separate the middle 61% of the distribution from the area in the tails of the standard normal distribution
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.
Middle 61%
Between the 50 - (61/2) = 19.5th percentile and the 50 + (61/2) = 80.5th percentile.
19.5th percentile.
Z with a pvalue of 0.195. So Z = -0.86
80.5th percentile.
Z with a pvalue of 0.805. So Z = 0.86.
Z-scores between -0.86 and 0.86 separate the middle 61% of the distribution from the area in the tails of the standard normal distribution
