$A$ tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of $1000$ cases.
Distance (in $km$) less than $4000$ $4000$ to $9000$ $9001$ to $14000$ more than $14000$ Frequency $20$ $210$ $325$ $445$
If you buy a tyre of this company,what is the probability that:
$(i)$ it will need to be replaced before it has covered $4000 \, km$?
$(ii)$ it will last more than $9000 \, km$?
$(iii)$ it will need to be replaced after it has covered somewhere between $4000 \, km$ and $14000 \, km$?
| Distance (in $km$) | less than $4000$ | $4000$ to $9000$ | $9001$ to $14000$ | more than $14000$ |
| Frequency | $20$ | $210$ | $325$ | $445$ |
If you buy a tyre of this company,what is the probability that:
$(i)$ it will need to be replaced before it has covered $4000 \, km$?
$(ii)$ it will last more than $9000 \, km$?
$(iii)$ it will need to be replaced after it has covered somewhere between $4000 \, km$ and $14000 \, km$?
(N/A) $(i)$ The total number of trials $= 1000$.
The frequency of a tyre that needs to be replaced before it covers $4000 \, km$ is $20$.
So,$P(\text{tyre to be replaced before it covers } 4000 \, km) = \frac{20}{1000} = 0.02$.
$(ii)$ The frequency of a tyre that will last more than $9000 \, km$ is the sum of frequencies for the ranges $9001$ to $14000$ and more than $14000$,which is $325 + 445 = 770$.
So,$P(\text{tyre will last more than } 9000 \, km) = \frac{770}{1000} = 0.77$.
$(iii)$ The frequency of a tyre that requires replacement between $4000 \, km$ and $14000 \, km$ is the sum of frequencies for the ranges $4000$ to $9000$ and $9001$ to $14000$,which is $210 + 325 = 535$.
So,$P(\text{tyre requiring replacement between } 4000 \, km \text{ and } 14000 \, km) = \frac{535}{1000} = 0.535$.
The frequency of a tyre that needs to be replaced before it covers $4000 \, km$ is $20$.
So,$P(\text{tyre to be replaced before it covers } 4000 \, km) = \frac{20}{1000} = 0.02$.
$(ii)$ The frequency of a tyre that will last more than $9000 \, km$ is the sum of frequencies for the ranges $9001$ to $14000$ and more than $14000$,which is $325 + 445 = 770$.
So,$P(\text{tyre will last more than } 9000 \, km) = \frac{770}{1000} = 0.77$.
$(iii)$ The frequency of a tyre that requires replacement between $4000 \, km$ and $14000 \, km$ is the sum of frequencies for the ranges $4000$ to $9000$ and $9001$ to $14000$,which is $210 + 325 = 535$.
So,$P(\text{tyre requiring replacement between } 4000 \, km \text{ and } 14000 \, km) = \frac{535}{1000} = 0.535$.
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