civil engineers belive that $W$, the amount of weight (in units of 1000 pounds) that a certain
span of a bridge can withstand without structural damage resulting, is normally distributed
with mean 400 and standard deviation 40. suppose that the weight (again,in units of 1000 pounds) of a car is a random variable with mean 3 and standard deviation .3.How many cars
would have to be on the bridge span for the probability of structural damage to exceed .1?
[Math] solving a problem in probability
probability
Related Solutions
If I recall correctly, you can take the values given to be the true population parameters.
You are not looking for $$P(6 < \bar X <8)$$ since this generally for a larger sample sizes, meaning $n\geq 2$. You are looking for $$P(6<X<8),$$ which I guess you could say $n = 1$.
You can check to see that the final answer is $0.3801842$.
You are correct that the answer is very nearly 0, but (even after @Drew's nice edit) it does not seem you are at all clear on how to get the answer.
Here is an outline:
1) The total weight $T = X_1 + X_2 + \cdots + X_{25}$ is normally distributed. Figure out its mean $\mu_T$ and its SD $\sigma_T.$
2) Standardize: $P(T > 6000) = P\left(Z = \frac{T-\mu_T}{\sigma_T} > \frac{6000 - \mu_T}{\sigma_T}\right).$
3) Evaluate the quantity $\frac{6000 - \mu_T}{\sigma_T}$, and then the probability, using standard normal tables or software.
Below is a graph of the density function of the distribution of $T$, Your answer is the area beneath the curve to the right of the vertical red line.
Best Answer
Make the obvious independence assumptions, and proceed along the lines described by Michael Chernick. Since everything has normal distribution, there is no need to make a normal approximation.
Let $X_i$ be the weight of the $i$-th car, and let $S=\sum X_i$. Then $S$ has normal distribution with mean $3n$ and variance $(0.3)^2 n$. Let $Y=S-W$. Then $Y$ has normal distribution with mean $3n-400$, and variance $v_n=(0.3)^2 n+(40)^2$.
We want to find smallest $n$ such that $\Pr(Y \gt 0) \gt 0.1$. To do that, we solve the equation $\Pr(Y \gt 0) = 0.1$. We have $$\Pr(Y\gt 0)=\Pr\left(\frac{Y-(3n-400)}{\sqrt{v_n}}\gt \frac{400-3n}{\sqrt{v_n}}\right)=\Pr\left(Z \gt \frac{400-3n}{\sqrt{v_n}}\right),$$ where $Z$ is standard normal.
But the point $c$ such that $\Pr(Z \gt c)=0.1$ is approximately given by $c=1.28$. So we need to solve the equation $$\frac{400-3n}{\sqrt{(0.3)^2 n+(40)^2}}=1.28.\tag{$1$}$$ Multiply through by the denominator, and square both sides. We get a quadratic equation in $n$, with somewhat messy coefficients. Solve as usual. We want the smaller root. (The larger root is extraneous, introduced when we squared both sides.)
Or else solve Equation $(1)$ directly by using the "Solve" button on a calculator, or any one of the many programs that do such things.
Our root is roughly $116.2$. So the number of cars such that the probability of damage is $\gt 0.1$ is $117$.
Remark: If we think about it, we can see that there is no need to solve a quadratic. For it is clear that the number of cars must be less than $400/3$. So the contribution of the cars to the variance of $Y$ is at most $(400/3)(0.09)$, which is $12$. The contribution of the bridge strength variance to the variance of $Y$ is $1600$, whose bulk makes $12$ negligible. Thus we can assume that $Y$ has standard deviation $40$, and for all practical purposes we want to solve the linear equation $$400-3n=(1.28)(40).$$