Classes are distributed between 49.0 and 54.0 minutes find probably that a given class period runs less than 50.75 minutes

Step-by-step explanation:

Assuming that this is a normal distribution, the mean is halfway between 49.0 and 54.0 minutes:  (49.0+54.0)/2 minutes = 51.5 minutes.

We want to find the area under the standard normal curve to the left of 50.75 minutes.  To do this, use the normalcdf( function on a TI-83 Plus calculator:

normalcdf(-1000,50.75,51.5, 1) = 0.23.

Notes:  The left boundary of the area in question is actually -infinity, but using -1000 as this boundary is only minimally different.  50.75 is the right area boundary.  51.5 is the mean.  In the absence of info from you regarding the standard deviation of this distribution, I have arbitrarily chosen to let the standard deviaton be 1.  Thus, the probability that a given class period runs less than 50.75 minutes is 0.23, if the standard deviation is 1.

If the standard deviation were 0.5 instead, then the probability in question would be

normalcdf(-1000,50.75,51.5, 0.5) = 0.31.

mr. graef told us it is 13 pints

step-by-step explanation: