Find the Z-scores that separate the middle 61% of the distribution from the area in the tails of the standard normal distribution

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

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

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?

i need more details sorry

step-by-step explanation: