Several science students used the same design when constructing a bridge out of balsa wood. They each tested the bridge’s weight capacity using cups filled with sand. The weight capacities were recorded, and the results follow a normal distribution curve. The mean of the weights is 19.6 lb with a standard deviation of 1.3 lb. Which weight is greater than 70% of the data?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(19.6,1.3)[/tex]

Where [tex]\mu=19.6[/tex] and [tex]\sigma=1.3[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.3[/tex] (a)

[tex]P(X<a)=0.7[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524. On this case P(Z<0.524)=0.7 and P(z>0.524)=0.3

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.7[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.7[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=0.524<\frac{a-19.6}{1.3}[/tex]

And if we solve for a we got

[tex]a=19.6 +0.524*1.3=20.28[/tex]

So the value of height that separates the bottom 70% of data from the top 30% is 20.28.