How do i solve this? : the times of all 15 year olds who run a certain race are approximately normally distributed with a given mean = 18 sec and standard deviation = 1.2 sec. what percentage of the runners have times less than 14.4 sec?

We need to 'standardise' the value of X = 14.4 by first calculating the z-score then look up on the z-table for the p-value (which is the probability)

The formula for z-score:
z = (X-μ) ÷ σ

Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2

Substitute these value into the formula

z-score = (14.4 - 18) ÷ 1..2 = -3

We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3)  but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)

From the table
P(Z < -3) = 0.0013

The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%

A=bxh3.2x1.8=5.76 centimeters square