# Thread: [SOLVED] Normal Distrubtion Question

1. ## [SOLVED] Normal Distrubtion Question

Question
The time taken by a garage to replace worn-out brake pads follows a normal distribution with mean 90 minutes and standard deviation 5.8 minutes. The garage claims to complete the replacement in 'a to b minutes'. If this claim is to be correct for 90% of the repairs, find a and b correct to 2 significant figures, based on a symmetrical intervals centred on the mean.

Attempt:
I only know that $\displaystyle \mu = 90$ and $\displaystyle \sigma = 5.8$. I can't understand what the question exactly means. How am I suppose to find the values of a and b ?

2. Originally Posted by looi76
Question
The time taken by a garage to replace worn-out brake pads follows a normal distribution with mean 90 minutes and standard deviation 5.8 minutes. The garage claims to complete the replacement in 'a to b minutes'. If this claim is to be correct for 90% of the repairs, find a and b correct to 2 significant figures, based on a symmetrical intervals centred on the mean.

Attempt:
I only know that $\displaystyle \mu = 90$ and $\displaystyle \sigma = 5.8$. I can't understand what the question exactly means. How am I suppose to find the values of a and b ?
Let $\displaystyle a = \mu - c$ and $\displaystyle b = \mu + c$.

Then $\displaystyle \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X > \mu + c) = 0.05$.

Alternatively, $\displaystyle \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X < \mu - c) = 0.05$.

Therefore $\displaystyle \mu - c = 80.4598$.

I leave the remaining details for you to finish.

3. Originally Posted by mr fantastic
Let $\displaystyle a = \mu - c$ and $\displaystyle b = \mu + c$.

Then $\displaystyle \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X > \mu + c) = 0.05$.

Alternatively, $\displaystyle \Pr(\mu - c < X < \mu + c) = 0.9 \Rightarrow \Pr(X < \mu - c) = 0.05$.

Therefore $\displaystyle \mu - c = 80.4598$.

I leave the remaining details for you to finish.
I should solve the problem using the normal distribution table. I have attached the table to this post.
For 90% the value is 1.282, why did you choose 1.645?

4. Originally Posted by looi76
I should solve the problem using the normal distribution table. I have attached the table to this post.
For 90% the value is 1.282, why did you choose 1.645?
The 10% (0.1) must be evenly divided between the two tails ..... 5% (0.05) each end .... So $\displaystyle \Pr(X > \mu + c) = 0.05 \, \Rightarrow \Pr (X < \mu + c) = 0.95$ ..... Therefore the corresponding z-value of $\displaystyle X = \mu + c$ is 1.645 .....

So $\displaystyle z = \frac{X - \mu}{\sigma} \Rightarrow 1.645 = \frac{c}{\sigma} \Rightarrow c = .....$