Question

Weights of 2-year-old girls are normally distributed with a mean of 253 lbs, and a standarddeviation of 1.12 lbs. According to this information, what weight would be the 33rd percentile? You must

Answer 1

We have here a case of a normally distributed variable. We can solve this kind of problem using the standard normal distribution, and the cumulative standard normal table (available in any Statistic Book, or on the internet).

We have that we can find z-scores to normalized the situation, and then, using the cumulative standard normal table, we can find the percentile. Then, we have:

`z=(x-\mu)/(\sigma)`

In this case, we need to find the raw value, x. We need to find a z-score that represents that before it there are 33% of the cases for this distribution: in this case, the value for z is approximately equal to z = -0.44.

Now, we have the mean (253 lbs), and the standard deviation (1.12 lbs):

`-0.44=(x-253)/(1.12)`

And now, we can determine the value, x, which is, approximately, the 33% percentile of this normal distribution:

1. Multiply by 1.12 to both sides of the equation:

`1.12\cdot(-0.44)=(1.12)/(1.12)\cdot(x-253)\Rightarrow-0.4928=x-253`

2. Add 253 to both sides of the equation:

`-0.4928+253=x-253+253\Rightarrow252.5072=x\Rightarrow x=252.5072`

Therefore, the weight that would be the 33rd percentile, is, approximately, x= 252.5072 or 252.51lbs (rounding to the nearest hundredth).