Extracted text: Suppose X is a random variable with X - N(90, 52) and let u = the mean of X me = the median of X mo = the mode of X a) P(X > 110) = P(X Extracted text: The middle 50% of the data will be from to Round your answers to 2 decimal places. d) For the measures of location of the random variable X it is true that те то.

Suppose X is a random variable with X - N(90, 52) and let u = the mean of X me = the median of X mo = the mode of X a) P(X > 110) = P(X

Suppose X is a random variable with<br>X - N(90, 52)<br>and let<br>u = the mean of X<br>me = the median of X<br>mo = the mode of X<br>a)<br>P(X > 110) = P(X <<br>).<br>b)<br>Calculate the percentage of the data that is within 1.4 standard deviations of the mean.<br>Round your answer to 2 decimal places.<br>

Extracted text: Suppose X is a random variable with X - N(90, 52) and let u = the mean of X me = the median of X mo = the mode of X a) P(X > 110) = P(X < ).="" b)="" calculate="" the="" percentage="" of="" the="" data="" that="" is="" within="" 1.4="" standard="" deviations="" of="" the="" mean.="" round="" your="" answer="" to="" 2="" decimal="">

The middle 50% of the data will be from<br>to<br>Round your answers to 2 decimal places.<br>d)<br>For the measures of location of the random variable X it is true that<br>те<br>то.<br>

Extracted text: The middle 50% of the data will be from to Round your answers to 2 decimal places. d) For the measures of location of the random variable X it is true that те то.

Jun 10, 2022

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Suppose X is a random variable with X - N(90, 52) and let u = the mean of X me = the median of X mo = the mode of X a) P(X > 110) = P(X

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