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Question: given the weight of a child is normally distributed with average weight = $25\,\text{kg}$ and variance = $4\,\text{kg}$. What is the probability that $7$ kids will weigh less than $180\,\text{kg}$?

My way of thinking is: $P(7K < 180)$. When converting this to the normal distribution, we need the expected value for 7 kids $= 7*25$ and the variance. By default, $var(a*X) = a^2var(X)$ for $a$ being a constant. However, for the normal distribution the variance of $7$ children is $7*var(X)$, without it being squared. Can someone explain to me what I am missing?

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Let $Z\sim N(\mu,\sigma^2)$ denote the normal distribution for the total weight of the $7$ children.

Now, we have:

$$Z=X_1+X_2+X_3+X_4+X_5+X_6+X_7$$ Where $X_k$ denotes the $k$-th child.

It is not $Z=7X$, since you are measuring $7$ different children's weights, and not multiplying the weight of one child by $7$ (i.e otherwise you would be assuming that all kids would weigh the same).

Now, you should get:

$$\operatorname*E(Z)=7\operatorname*E(X)$$ And $$\operatorname*{Var}(Z)=\operatorname*{Var}(X_1)+\operatorname*{Var}(X_2)+\cdots+\operatorname*{Var}(X_7)=7\operatorname*{Var}(X)$$

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$7K=K+K+K+K+K+K+K$ would mean that all kid weigh the same, which is $K$, that is not correct.

The weight of kid follows a normal distribution. Let $X_k$ the weight of the $k^{th}$ kid, we can add that the weight between the kids are independent. What you want to compute is $$P(\sum_{k=1}^{7}X_k < 180)$$

The random variable S defined as $S=\sum_{k=1}^{7}X_k $ is normally distributed being the sum of independent normal variables. Therefore $S$ is entirely defined by its mean and its variance.

We know that $$E(S)=\sum_{k=1}^{7}E(X_k)=7*25=175$$

and $$var(S)=var(\sum_{k=1}^{7}X_k)=\sum_{k=1}^{7}var(X_k)=4*7=28$$

Finally,

$$P(\sum_{k=1}^{7}X_k < 180)=P(175+\sqrt{28}Y< 180)=P(Y<\frac{5}{\sqrt{28}})=\Phi(\frac{5}{\sqrt{28}}) \approx 82.8 \%$$

where $Y$ is a standard normal variable, and $\Phi$ its cdf.

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  • $\begingroup$ My mind is actually blown, did this wrong during the entire course, always writing it down as k*X in stead of X + X + ... Thank you so much, you saved tomorrow's exam :) $\endgroup$ – Lander Jan 18 '17 at 21:18
  • $\begingroup$ You meant instead of $$X_1+X_2+...$$ $\endgroup$ – Canardini Jan 18 '17 at 21:19

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