Respuesta :

51 inches is 2 inches more than the mean, which is 1 standard deviation.

Using the 68, 95, 99.7 rule:

100-68 / 2 = 16% would be taller than 51 inches.

The approximate percentage of children aged 7-years who are taller than 51 inches is 16%.

What is a Z-table?

A z-table also known as the standard normal distribution table, helps us to know the percentage of values that are below (or to the left of the Distribution) a z-score in the standard normal distribution.

As it is given that the mean of the children's height is 49 inches, while the standard deviation is 2 inches. Therefore, the approximate percentage of children aged 7-years who are taller than 51 inches can be written as,

[tex]P(X > 51) = 1-P(x \leq 51)[/tex]

                  [tex]= 1 - P(z\leq \dfrac{X-\mu}{\sigma})\\\\= 1 - P(z\leq \dfrac{51-49}{2})\\\\\text{Using the z-table}\\\\=1-0.8413\\\\= 0.1587\approx 0.16[/tex]

Thus, the approximate percentage of children aged 7-years who are taller than 51 inches is 16%.

Learn more about Z-table:

https://brainly.com/question/6096474