Asiah found that the time taken to travel from her house to her office is normally distributed with a mean of 40 minutes and a varience of 64 minutes^2.If she leaves home at 7:30am,Find the probability that she will arrive after 8:15 am
please help
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Asiah found that the time taken to travel from her house to her office is normally distributed with a mean of 40 minutes and a varience of 64 minutes^2.If she leaves home at 7:30am,Find the probability that she will arrive after 8:15 am
please help
Transform it to the standard $\displaystyle Z$ distribution and look it up on your normal probability tables...
$\displaystyle \sigma^2=64\Rightarrow\sigma=+\sqrt{64}=8\;\text{\ footnotesize\ minutes}$Quote:
Originally Posted by mastermin346
You want the probability that the time taken is >45 minutes...
therefore, you require
$\displaystyle P\left(Z>\frac{x-\mu}{\sigma}\Rightarrow\ Z>\frac{45-40}{8}\right)$
thanks archie meade
how about this..
In standard sports of a school,all students took part in the 100m race.The time taken to finish the race by a Form 2 student in normally distributed with a mean of 15 seconds and variance of 25 seconds^2.For a student who took less than 16 seconds,1 mark was awarded and for one who took less than 14 seconds,2 marks were awarded.
a)find the percentage of Form 2 students who got 1 mark each.
b)Find the number of Form 2 students who got 2 marks if 200 Form 2 students took part.
(a)Quote:
Originally Posted by mastermin346
Again, the question gives the variance, so the standard deviation is the square root of that...
$\displaystyle \sigma=+\sqrt{25}=5$
$\displaystyle \mu=15$
$\displaystyle Z=\displaystyle\frac{x-\mu}{\sigma}=\frac{16-15}{5}=0.2$
Therefore, you need to find
$\displaystyle P(Z<0.2)$
The percentage of students who got one mark is then 100 times the probability
of a student finishing the race in less than 16 seconds.
Try that and then maybe (b) is do-able.
