The lengths of pregnancies in a small rural village are normally distributed with a mean of 264.1 days and a standard deviation of 12.9 days. In what range would you expect to find the middle 95% of most pregnancies

The range of the 95% data (X) =  238.3 days <  X < 289.9 days

Step-by-step explanation:

Given;

mean of the normal distribution, m = 264.1 days

standard deviation, d = 12.9 days

between two standard deviation below and above the mean is 96% of all the data.

two standard deviation below the mean = m - 2d

                                                                   = 264.1 - 2(12.9)

                                                                   = 238.3 days

two standard deviation above the mean = m + 2d

                                                                    = 264.1 + 2(12.9)

                                                                     = 289.9 days

The middle of the 95% of most pregnancies would be found in the following range;

238.3 days <  X < 289.9 days

since both sides equal, there are infinitely many olutions.

it could be 1,3 and more others to create a cylinder.

step-by-step explanation: