A set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. Find the value that is –2 standard devi
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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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