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1zephyr · Sep 2, 2015

I dunno where i can post this but you guys are good at maths

A car manufacturer takes an average of 17.5 hours to construct a car. This includes time for stamping, welding, painting, assembly and inspections. Construction times vary with a standard deviation of 30 minutes and these times follow a normal distribution

b. What is the probability that a randomly selected car manufactured at this plant takes between 18 and 19 hours to construct?

c. Find the probability that the construction time for a randomly selected car manufactured at this plant is less than 17 hours and 50 minutes.

Pretty sure ive done it wrong can anyone help?

InteGrand · Sep 2, 2015

1zephyr said:
I dunno where i can post this but you guys are good at maths

A car manufacturer takes an average of 17.5 hours to construct a car. This includes time for stamping, welding, painting, assembly and inspections. Construction times vary with a standard deviation of 30 minutes and these times follow a normal distribution

b. What is the probability that a randomly selected car manufactured at this plant takes between 18 and 19 hours to construct?

c. Find the probability that the construction time for a randomly selected car manufactured at this plant is less than 17 hours and 50 minutes.

Pretty sure ive done it wrong can anyone help?
View attachment 32352 View attachment 32351

I think you've done it right.

$If $T$ is the construction time in hours, we are given that $T \sim \mathcal{N}\left( \mu,\sigma^2 \right )$, where $\mu = \mathbb{E}[T] = 17.5$, and $\sigma^2 = \mathbb{V}\mathrm{ar}[T]=(0.5)^2=0.25$ (since the standard deviation is $\sigma =0.5$ hours).$

$This can be transformed to a standard normal random variable by writing $Z=\frac{T-\mu}{\sigma}$, so $Z\sim \mathcal{N}(0,1)$.$

$b) We want to find $\mathbb{P}\left(18<T<19 \right)$. We have$

$$\mathbb{P}\left(18<T<19 \right) = \mathbb{P}\left(\frac{18-\mu}{\sigma}<Z <\frac{19-\mu}{\sigma}\right)$ (standard normal transformation)$

$$= \mathbb{P}\left(\frac{18-17.5}{0.5}<Z <\frac{19-17.5}{0.5}\right)$ (substituting given values for $\mu$ and $\sigma$)$

$= \mathbb{P}\left(1<Z <3\right)$

$$= \Phi (3) - \Phi (1)$, where $\Phi (z)=\int _{-\infty} ^{z} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}t^2}\text{ d}t$ is the standard normal CDF$

$$\approx 0.9987 - 0.8413$ (using values from a standard normal probability table, e.g. from)$ http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/Statistics/Z-Table.pdf

$$=0.1574$ (answer).$

$c) We know that 17 hours and 50 minutes is $17\tfrac{50}{60}=17.8\dot{3}$ hours. So we want to find $\mathbb{P}\left(0<T< 17.8\dot{3}\right)$. We have$

$\mathbb{P}\left(0<T< 17.8\dot{3}\right)=\mathbb{P}\left(\frac{0-\mu}{\sigma}<Z< \frac{17.8\dot{3}-\mu}{\sigma}\right)$

$=\mathbb{P}\left(\frac{0-17.5}{0.5}<Z< \frac{17.8\dot{3}-17.5}{0.5}\right)$

$=\mathbb{P}\left(-35<Z< 0.\dot{6}\right)$

$=\Phi\left(0.\dot{6}\right)-\Phi \left(-35 \right)$

$$\approx \Phi\left(0.\dot{6}\right)$, since $\Phi \left(-35 \right)$ is basically 0$

$$\approx \frac{1}{3}\Phi \left(0.66 \right)+\frac{2}{3}\Phi \left(0.67 \right)$ (using linear interpolation between two values which are on the standard normal CDF table)$

$\approx \frac{1}{3}\times 0.7454 +\frac{2}{3}\times 0.7486$

$=0.7475\dot{3}$

$$\approx 0.748$.$

1zephyr · Sep 2, 2015

ok sweet
thanks for taking the time to show us your working

1zephyr · Sep 2, 2015

follow up
Find the construction time of a car which is on the 10th percentile of this distribution.
I need to draw a probability distribution plot but i dont know what to put in the mean and standard deviation as the results i get dont seem right
Can anyone possibly make a graph

InteGrand · Sep 2, 2015

1zephyr said:
follow up
Find the construction time of a car which is on the 10th percentile of this distribution.
I need to draw a probability distribution plot but i dont know what to put in the mean and standard deviation as the results i get dont seem right
Can anyone possibly make a graph

Here's a link with the plot of the PDF, as well as the CDF and values for some percentiles:

http://www.wolframalpha.com/input/?i=normal+distribution+mean+17.5,+standard+deviation+0.5

1zephyr · Sep 2, 2015

InteGrand said:
Here's a link with the plot of the PDF, as well as the CDF and values for some percentiles:

http://www.wolframalpha.com/input/?i=normal+distribution+mean+17.5,+standard+deviation+0.5

ahh youre a lifesaver, thank you so much ive been doing stuff like putting 30 for standard deviation or 1050 minutes for the mean

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