Babies were weighed just after being born and here are the results:

Mean weight: 5.33 lbs, Standard Deviation: 0.65, A baby severely underweight classified by the hospital is 4.85 lbs

The hospital wants to take a picture of the 10% heaviest babies born in that hospital. What would be the lightest weight to qualify for a photo?

Using the normalcdf function, I believe the number of underweight babies is 0.23. Not even sure if that's correct.


closed as off-topic by mdewey, Sycorax, gung, Tim, Kodiologist Oct 18 '16 at 19:02

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  • 3
    $\begingroup$ Please consider adding the tag self-study and follow the guidelines after reading its wiki. $\endgroup$ – Firebug Oct 17 '16 at 20:07
  • $\begingroup$ What are you talking about? $\endgroup$ – Kimberly Oct 17 '16 at 20:14
  • $\begingroup$ Do you want to post your code or explain what your steps are? $\endgroup$ – ilanman Oct 17 '16 at 20:20
  • 1
    $\begingroup$ You are assuming a normal distribution as a given. That has to be stated explicitly or the question has no answer. $\endgroup$ – Carl Oct 17 '16 at 20:26
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    $\begingroup$ @Kimberly It's stats.stackexchange.com policy to ask posters to add the self-study tag to questions which are related to textbook exercises, homework etc... self-study tag signals that answers & comments should improve understanding, point you in the right direction, help you solve the problem yourself. $\endgroup$ – Matthew Gunn Oct 17 '16 at 20:38

Let $X$ be a random variable denoting a baby's weight and let $F$ be the cumulative distribution function (CDF) for $X$. Recall that the CDF gives:

$$ F(x) = P(X \leq x) $$

You want to find a weight $x$ such that $90\%$ of babies weigh that amount or less and $10\%$ of babies weigh more. That is, you want to find an $x$ such that:

$$ .9 = F(x) $$

To proceed you would need to know the distribution of $X$ (i.e. you would need to know what the CDF $F$ is). Not everything automatically follows the normal distribution!

A useful fact for the normal distribution is that the CDF for a normally distributed random variable with mean $\mu$ and standard deviation $\sigma$ can be written as:

$$F(x\, ; \mu, \sigma) = \Phi\left( \frac{x - \mu}{\sigma} \right)$$

where $\Phi(x)$ is the CDF for a standard, normally distributed random variable (i.e. with mean 0 and standard deviation 1).

  • $\begingroup$ Thank you, I was confused what normal distributed random variable was. It was not stated in my question but I'm assuming according to the previous homework it was that the mean is 0 and standard deviation is 1. $\endgroup$ – Kimberly Oct 17 '16 at 21:09
  • $\begingroup$ @Kimberly A normally distributed random variable has a distribution that looks like a bell curve. The normal distribution has two parameters: the mean $\mu$ and the standard deviation $\sigma$. The special case where $\mu=0$ and $\sigma =1$ is called the standard normal distribution. $\endgroup$ – Matthew Gunn Oct 17 '16 at 21:13
  • $\begingroup$ Got it thank you! That's a lot more helpful than my professor. So knowing this I'm assuming I will be using the invnorm function? $\endgroup$ – Kimberly Oct 17 '16 at 21:14
  • $\begingroup$ @Kimberly Yes. invnorm is probably $\Phi^{-1}$ (I don't know your system). So $x = \Phi^{-1}(.9) \sigma + \mu $ (of course assuming $X$ is normally distributed.) $\endgroup$ – Matthew Gunn Oct 17 '16 at 21:16
  • $\begingroup$ @Kimberly the invnorm command may even allow you to pass in $\mu$ and $\sigma$ (i.e. give you the inverse CDF for a normally distributed random variable with a specific $\mu$ and $\sigma$. $\endgroup$ – Matthew Gunn Oct 17 '16 at 21:17

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