178k views
4 votes
How to solve question 19
thanks

How to solve question 19 thanks-example-1

1 Answer

3 votes
Let
X be the random variable for the weight of any given can, and let
\mu and
\sigma be the mean and standard deviation, respectively, for the distribution of
X.

You have


\begin{cases}\mathbb P(X<347)=0.0015\\\mathbb P(X>377)=0.025\end{cases}

Recall that for any normal distribution, approximately 99.7% of it lies within three standard deviations of the mean, i.e.
\mathbb P(\mu-3\sigma<X<\mu+3\sigma)=0.997. This means 0.3% must lie outside this range,
\mathbb P(X<\mu-3\sigma~\lor~X>\mu+3\sigma)=0.003. Because the distribution is symmetric, it follows that
\mathbb P(X<\mu-3\sigma)=0.0015.

Also recall that for any normal distribution, about 95% of it falls within two standard deviations of the mean, so
\mathbb P(\mu-2\sigma<X<\mu+2\sigma)=0.95, which means 5% falls outside, and by symmetry,
\mathbb P(X>\mu+2\sigma)=0.025.

Together this means


\begin{cases}\mu-3\sigma=347\\\mu+2\sigma=377\end{cases}

Solving for the mean and standard deviation gives
\mu=365 and
\sigma=6.
User Katiria
by
7.4k points

No related questions found