It is given that we have a normal distribution, the mean is given as 105 centimeters and the standard deviation is 5 centimeters.

The z-score for 114 centimeters is z_{1} = \frac{114-105}{5}= 1.8

The z-score for 118 centimeters is z_{2} = \frac{118-105}{5} = 2.6

Given below is the normal distribution curve and we need to find the area which is highlighted or the proportion of temperatures between these two z-scores:

Looking up z_{1} = 1.8 in the z-score table we see that 0.9641 of student heights are below 114 cm.

Looking up z_{2} = 2.6 in the z-score table we see that 0.9953 of student heights are below 118 cm.

To find the highlighted area we need to subtract the area below z_{1} from the area below z_{2}

So, 0.9953 - 0.9641 = 0.0312.

So, 0.0312 proportion of student heights are between 114 centimeters and 118 centimeters.

I found an answer from statweb.stanford.edu

Eric Min

The **definition** of a linear model goes further than a straight line. ... A **standard**
normal **distribution** is Z ∼ N(0,1) and is characterized by the following (where z ...

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