MHB Estimate the cost for tyres using the mean

[mathlearn]

Data

Data obtained from a survey conducted by a tyre manufacturing company, using a sample of 60 tyres of a certain category, on the distance tyres ran before they became unsuitable for further use.


View attachment 5986Problem



i. What is the modal class

ii. Taking the mid value of the modal class as the assumed mean , or otherwise , FInd the mean running distance of the tyres

iii. A businessman who uses tyres of the above category for his 10 motor cars each having four wheels , needs to estimate the annual cost for tyres. Assuming that every motor car runs 25,000 km a year on average and the price of one tyre is $ 10,000 estimate the cost.

What has already been done by myself:

Found out the modal class which is 36-42.

I calculated the assumed mean.

Class interval	Mid value(x)	Deviation(d)	frequency(f)	fd
18-24	30	-27	1	-27
24-30	39	-18	5	-90
30-36	48	-9	6	-54
36-42	57	0	26	0
42-48	66	9	15	135
48-54	75	18	7	126
			Σf=60	Σfd=90

$Mean=Assumed mean+\frac{Σfd}{Σf}$

$Mean=57+\frac{90}{60}$
$Mean=57+\frac{2}{3}$
$Mean=58\frac{1}{2}$Where do I need help

Thinking that I have correctly found the mean, to build that argument in iii.

iii. A businessman who uses tyres of the above category for his 10 motor cars each having four wheels , needs to estimate the annual cost for tyres. Assuming that every motor car runs 25,000 km a year on average and the price of one tyre is $ 10,000 estimate the cost.

Many Thanks :)

 

[I like Serena]

iii. A businessman who uses tyres of the above category for his 10 motor cars each having four wheels , needs to estimate the annual cost for tyres. Assuming that every motor car runs 25,000 km a year on average and the price of one tyre is $ 10,000 estimate the cost.

Hey mathlearn! ;)

Just putting all the information next to each other, we need:
10 motor cars times 4 tires times 25,000 km a year.

We know each tire runs 58,500 km on average before running out.
So dividing by that number tells us how many tires we need per year (on average).

Multiply by the cost per tire to get the estimate of the yearly cost.

 

[mathlearn]

Hey mathlearn! ;)

Just putting all the information next to each other, we need:
10 motor cars times 4 tires times 25,000 km a year.

We know each tire runs 58,500 km on average before running out.
So dividing by that number tells us how many tires we need per year (on average).

Multiply by the cost per tire to get the estimate of the yearly cost.

So the distance run by the 4 tyres of each car are equal.

As the distance run by a tyre before worn out is greater than the distance expected by the cars to run , The tyres which are already used in the cars would be sufficient.

Then the cost of tyres would be 4*10*10,000=400,000 $

Correct? :)

 

[I like Serena]

So the distance run by the 4 tyres of each car are equal.

As the distance run by a tyre before worn out is greater than the distance expected by the cars to run , The tyres which are already used in the cars would be sufficient.

Then the cost of tyres would be 4*10*10,000=400,000 $

Correct? :)

That's correct to find the initial purchasing cost.
However, the question asks for the "annual cost" instead.
That assumes we'll be using them for many years, and we want to know what it will take on average in anyone year.

 

[mathlearn]

That's correct to find the initial purchasing cost.
However, the question asks for the "annual cost" instead.
That assumes we'll be using them for many years, and we want to know what it will take on average in anyone year.

So how many years should be considered ? (Happy)

 

[I like Serena]

So how many years should be considered ? (Happy)

It doesn't say - we can treat it as an infinite number of years, effectively making the initial purchasing cost irrelevant. (Thinking)

 

[mathlearn]

Using an online calculator I get the mean value as 37. Not sure of it's accuracy.

View attachment 5989

Still does it make any difference?

 

[I like Serena]

Using an online calculator I get the mean value as 37.

Still does it make any difference?

We didn't use the mean yet... (Thinking)

Suppose we look at just 1 tire that runs 25000 km each year.
Then we can expect that we have to replace it after 37000 / 25000 years at which time it will cost $10000 (is a tire really that expensive? (Giggle)).
It means that the annual cost of 1 tire is $10000 x 25000 / 37000.

 

[I like Serena]

Using an online calculator I get the mean value as 37. Not sure of it's accuracy.

Let's take a look at the first row.
I see we have the category 18:24 with a frequency of 1.
It has listed that Freq*X = 18.
Shouldn't that be 21, which is the mid point of the category? (Wondering)

 

[mathlearn]

Let's take a look at the first row.
I see we have the category 18:24 with a frequency of 1.
It has listed that Freq*X = 18.
Shouldn't that be 21, which is the mid point of the category? (Wondering)

Yes , well then the calculator is wrong (Nod)

We didn't use the mean yet... (Thinking)

Suppose we look at just 1 tire that runs 25000 km each year.
Then we can expect that we have to replace it after 37000 / 25000 years at which time it will cost $10000 (is a tire really that expensive? (Giggle)).
It means that the annual cost of 1 tire is $10000 x 25000 / 37000.

Thank you very much (Yes) , The denominator of the fraction should be replaced with the mean calculated in the first post. And the tyre is expensive (Giggle)