Using the normal distribution, it is found that there was a 0.9579 = 95.79% probability of a month having a PCE between $575 and $790.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation are given, respectively, by:
.
The probability of a month having a PCE between $575 and $790 is the <u>p-value of Z when X = 790 subtracted by the p-value of Z when X = 575</u>, hence:
X = 790:


Z = 1.8
Z = 1.8 has a p-value of 0.9641.
X = 575:


Z = -2.5
Z = -2.5 has a p-value of 0.0062.
0.9641 - 0.0062 = 0.9579.
0.9579 = 95.79% probability of a month having a PCE between $575 and $790.
More can be learned about the normal distribution at brainly.com/question/4079902
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Z for the 5th percentile is about -1.6449
1,000,000 is 1 million
look at the previous digit
if it is less than 5 then leave the millions place as is and replace the following digits with zeroes
if it is equal to or greater than 5, add 1 to the millions place and replace the following digits with zeroes
296,386,482
3<5
round down (leave as is baseically)
296,000,000 is rounded
(4x+3) (2x-1)=0
X= -3/4, 1/2