0
$\begingroup$

The lifespan of a certain brand of bicycle tyre is normally distributed witha mean of 1200km and a std deviation of 80km.

1)Find the probability that a randomly chosen tyre has a lifespan of less than 1100km.

2)Determine the value of A, the nearest integer , if it is found that 4% of the tyres are not road-worthy after A km

For 1), we just use P(X<1100) whr ans is 0.1056, is it?

For 2), how do we do it?

1 Answers 1

0

For the first part you find out P(X<1100) by converting it into standard distribution form and then looking up the value of integration from -inf to obtained value from cumulative normal distribution table. For second case you are given the probability as .04. Look up from the table the value of standardised lifespan A and then scale it to given mean and SD. So the second part is just inverse path of first part.

  • 0
    sorry, i didnt learn "integration form -inf" ;; – 2017-02-16
  • 0
    You don't have to integrate. You have P((X-u)/sd < (1100-1200)/80). You need to look up the value (-100/80 )= -1.25 in a standard normal distribution table ( https://f.hypotheses.org/wp-content/blogs.dir/253/files/2013/10/Capture-d’écran-2013-10-15-à-14.22.40.png) . Notice that it is a table of integration from -inf to X. – 2017-02-16
  • 0
    Ahhh I see I see, so for first part, it's just basic. What about 2nd part? I'm still not rly sure, is it 0.04 = (x-1200)/80, then find the X? – 2017-02-16
  • 0
    No, second part would be, 0.04 = P( Z < (A - 1200)/80). Now use the table in opposite manner to find A. – 2017-02-16
  • 0
    Ah Icic, thanks alot ! – 2017-02-16