# basic standard deviation question

The weights of newborn children in the United States vary according to the normal distribution with mean 7.5 pounds and standard deviation 1.25 pounds. The government classifies a newborn as having low birth weight if the weight is less than 5.5 pounds.

b) What weight do only 1% of U.S. newborns exceed?

My work: looked at a table to find the Z score of 1 percent which was 3.49 (using the table given in the basic practice for statistics)

then using the equation z = x - mean/standard deviation - 3.49(1.25) + 7.5 = 11.9

But the answer is 10.40793485 lb, not sure what i did wrong and thanks for the help.

## 2 Answers

As commented, the $z$-score value is strange. Since we want to find the $1\%$ of newborns who exceed a certain weight, we can conduct a $99\%$ one-sided confidence interval to determine this.

I used a $z$-score of $2.327$ ("guesstimate") for the calculation.

If you look at the normal table you will find that $\Pr(Z\gt 2.33)\approx 0.01$, or equivalently $\Pr(Z\le 2.33)\approx 0.99$. (The right number is between $2.32$ and $2.33$.)

It looks as if there was in error in the table look up.

• now i feel stupid i was looking at 100 percent not 99 – cmptUser Jun 26 '16 at 0:15
• @RishiDholliwar: In principle, there is no place $z$ such that $\Pr(Z\le z)=100\%$. But indeed $\Pr(Z\gt 3.5)$ is $0$ for most practical purposes. – André Nicolas Jun 26 '16 at 0:22
• @RishiDholliwar: Side comment about the "answer" of $10.4079\dots$. That kind of precision is very unreasonable. The mean is only given to $2$ significant figures, and the standard deviation to $3$. And perhaps the weight distribution is roughly normal, but it is not clear how good an approximation that is. – André Nicolas Jun 26 '16 at 4:49