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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.

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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.

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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.

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  • $\begingroup$ now i feel stupid i was looking at 100 percent not 99 $\endgroup$ – cmptUser Jun 26 '16 at 0:15
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Jun 26 '16 at 0:22
  • $\begingroup$ @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. $\endgroup$ – André Nicolas Jun 26 '16 at 4:49

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