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

beanburito

step-by-step explanation:

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: