Question

Can anyone help me with binomial distribution please??? Im very confused There are 3 questions i will type the first one

Answer 1

A company maufactures batteries having life spans that are normally distributed, with a mean of 45 mo and a standard deviation of 5 mo find the probability that the battery chosen at randomm will have a life span of 50-55 mo

Answer 2

I know that p is 13.6 because i used a graphing calc but im supposed to show work?

Answer 3

@dan815 ?? Can you help?

Answer 4

@Nnesha

Answer 5

You're given that the life span of the batteries is normally distributed. What help would the binomial distribution provide here?

Anyway, denote the lifespan of any given battery by \(X\). You're asked to find \(P(50<X<55)\). You know that the mean is \(45\) and the standard deviation is \(5\). Transform to the standard normal distribution using the formula \(Z=\dfrac{X-\mu}{\sigma}\).
\[\begin{align*}P(50<X<55)&=P\left(\frac{50-45}{5}<\frac{X-45}{5}<\frac{55-45}{5}\right)\\
&=P(1<Z<2)\\
&=P(Z<2)-P(Z<1)
\end{align*}\]
From here, you can use a table of \(z\) scores to find the left-tail probabilities or, if you're not familiar with this way of computing probabilities, you can use the standard deviation rule.
http://en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule

Answer 6

@SithsAndGiggles thank you for answering but I'm not sure what it 2,1 is supposed to be? i calculated that the probability was around 13.6 but is that wrong

Answer 7

\(P(1<Z<2)\) translates to the area under the curve for values of \(z\) that lie between 1 and 2 standard deviations from the mean.