1
$\begingroup$

I'm having trouble picking out what is what in this problem.

"The weight of people in a small town in Missouri is known to be normally distributed with a mean of $187$ pounds and a standard deviation of $29$ pounds. On a raft that takes people across the river, a sign states, “Maximum capacity $3,468$ pounds or $17$ persons.” What is the probability that a random sample of $17$ persons will exceed the weight limit of $3,468$ pounds?"

I think that $\mu=187$, $\sigma=29$, $n=17$, and $\bar{x}=3468$; I think the calculation I'm supposed to use is $z= \sqrt{17}(\bar{x} - \mu) / \sigma$. but I keep getting a value that is WAY too large for the z table.

Can someone please explain what I'm doing wrong?

$\endgroup$
1
  • $\begingroup$ is this is HW please add the self-study tag $\endgroup$
    – Antoine
    Mar 1, 2016 at 15:49

2 Answers 2

2
$\begingroup$

If each person has a weight that is normal$(\mu = 187, \sigma^2 = 841)$ then assuming independence across the samples (we won't have perfect independence because the sampling is without replacement but it's close enough) then the total weight of $17$ randomly selected people is $\sum_{j=1}^{17} X_j \sim$ normal$(\mu = 3179, \sigma^2 = 14297)$ (we have added both the means and variances for each $X_j$). The probability we want to compute is

\begin{align} P \left ( \sum_{j=1}^{17} X_j > 3468 \right ) &= P \left ( \frac{\sum_{j=1}^{17} X_j - 3179}{\sqrt{14297}} > \frac{3468 - 3179}{\sqrt{14297}} \right ) \\ &= P(Z > 2.413) \end{align}

where $Z \sim$ normal$(\mu = 0, \sigma^2 = 1)$. This probability can be determined using either a table or software.

$\endgroup$
0
$\begingroup$

There are two ways to solve the problem: Either by formulating your z-statistic in terms of the sum of normal variabels, or based on the average average value of the normal variables.

Your problem begins at the latter appproach, but there is one step that is not correct. You have specified that $\bar{x}=3468$, indicating that the maximal average weight of the persons on the raft is 3.4 thousand pounds before it sinks (the bar normally denotes average). In order to use the proceduree you have chosen, $\bar{x}$ should be the maximal average weight of 17 persons on the raft (meaning their total weight should not exceed 3468.

The maximum average weight is then $3468/17=204$ pounds. Inserting this into your z-statistic gives:

$Z=\frac{\sqrt{17}(204-187)}{29}=2.41$

The associated p-value can be found in a table where $P(sinks)=1-P(Z<2.41)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.