Question

In an 'eight' rowing boat there are eight rowers and a coxswain, who gives instructions to the rowers. The distribution of masses, in kg, of rowers is given by

M ~ N (92.5, 1 . 8^{2})

. The distribution of the mass, in kg, of the coxswain is given by

C ~ N (57, 0 . 6^{2}) .

Assuming that the masses of the rows are independent, find the probability that in a randomly chosen 'eight' rowing boat the total mass of the rowers and coxswain is less than

800 kg

.

Accepted Answer

Given data:

In an 'eight' rowing boat there are eight rowers and a coxswain, who gives instructions to the rowers. The distribution of masses, in kg, of rowers is given by $M ~ N (92.5, 1 . 8^{2})$ . The distribution of the mass, in kg, of the coxswain is given by $C ~ N (57, 0 . 6^{2}) .$ Assuming that the masses of the rows are independent,

Also given that total mass of the rowers and coxswain is less than $800 kg$ .

For $X ~ N (μ, σ^{2})$ indicates that $X$ is a Normal random variable, where $μ = E (X)$ and $σ^{2} = Var (X)$ and standard deviation $= \sqrt{Variance}$ .

The random variable here is the sum of nine independent observations, which are the masses of rowers and the mass of coxswain, i.e., $(M_{1} + M_{2} + \dots + M_{8} + C)$ .

We know that, if the random variable is the sum of $n$ independent observations of $X$ , i.e., $X_{1} + X_{2} + \dots + X_{n}$ then $Mean (X_{1} + X_{2} + . . . . . X_{n}) = n \times Mean (X)$ and $Var (X_{1} + X_{2} + . . . . . X_{n}) = Var (X_{1}) + Var (X_{2}) + . . . . . . + Var (X_{n})$ and $Mean (a X + b Y) = a \times Mean (X) + b \times Mean (Y)$ , $Var (a X + b Y) = a^{2} \times Var (X) + b^{2} \times Var (Y)$ .

Now, $Mean (M_{1} + M_{2} + \dots + M_{8} + C) = 8 \times 92.5 + 57 = 797$

$Var (M_{1} + M_{2} + \dots + M_{8} + C) = 8 \times 1 . 8^{2} + 0 . 6^{2} = 26.28$

If two or more independent normal distributions are combined, the resulting distribution is also normal.

$\Rightarrow M_{1} + M_{2} + \dots + M_{8} + C ~ N (797, 26.28)$

We know that, $P (X < n) = P (z < \frac{n - μ}{\sqrt{σ^{2}}})$ , where $n$ is observed value.

$\Rightarrow P (M_{1} + M_{2} + \dots + M_{s} + C < 800) = P (z < \frac{800 - 797}{\sqrt{26.28}})$

$= P (z < 0.58)$

From the $z$ -scores table,

$= 0.7190$

$= 0.281$

Therefore, $P (M_{1} + M_{2} + \dots + M_{8} + C < 800) = 0.7190$ .

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