Probability that the weight of the carts exceeds 255 tons

[Addez123]

Summary:: When filling up carts with iron the real weight deviates from the nominal value 10 ton. The standard deviation is .5.
What's the probability that 25 carts exceed 255 ton?

The arithmetic median value is:
$$X \in N(25 * 10, 0.5 / \sqrt{25}) = N(250, .1)$$
$$P(x > 255) = 1 - P(x < 255) $$
$$1 - Φ( \frac {255 - 250} {.1} ) = 1 - Φ(50) \approx 0$$

The book says answer is 1 - Φ(2)
but i see no way of that being true.

 

[etotheipi]

Using the notation $X \sim N(\mu, \sigma^2)$, each cart is distributed like$$C \sim N(10, \frac{1}{4})$$Then the sum of 25 carts is distributed like$$C_T = C_1 + C_2 + \dots + C_{25} \sim N(250, \frac{25}{4})$$Now you just require $P(C_T > 255)$ in terms of the function $\Phi$. That is just a matter of working out the $z$-score of a value $255$ using the correct variance.

 

[hutchphd]

When you have multiple independent random trials, the total grows like $N$ and the SD grows like $\sqrt N$. It is the percent error (CV) that diminishes like $\sqrt N$
So $\sigma =2.5$

 

[Addez123]

I know what I did wrong.
My formula is to calculate the median value (even though the formula was incorrect aswell), when I should be using addition of normal distribution.
$$N(25 * 10, .5 * \sqrt{25})$$