Normal Distribution (gas consumption in vehicles)

[TheNextOne]

Hi all,

I was wondering if anybody could help me with the following questions

Suppose gas consumption for a particular type of car follows a normal distribution with mean 15 kilometres per litre and a standard deviation 4 kilometres per litre.

a) Four cars of this type are selected randomly. The probability that 3 of them will have gas consumption over 16 kilometres per litre is closest to ?

b) Four hundred (400) cars of this type are selected randomly. The probability that at least 220 of them will have gas consumption over 15 kilometres per litre is closest to ?

This is what I have done:

a) (4 choose 3) P(x>16)^3 (1-P(x>16))^1

I dont know what P (X>16) is.

b) smaple mean= 200
sample standard deviation = 10

Thus, P(x is greater than or equal to 220)= (219.5-200)/10
= 1.95

And then would I find the corresponding z-score value?

 

[Guest]

Hi all,

I was wondering if anybody could help me with the following questions

Suppose gas consumption for a particular type of car follows a normal distribution with mean 15 kilometres per litre and a standard deviation 4 kilometres per litre.

a) Four cars of this type are selected randomly. The probability that 3 of them will have gas consumption over 16 kilometres per litre is closest to ?

This is what I have done:

a) (4 choose 3) P(x>16)^3 (1-P(x>16))^1

I dont know what P (X>16) is.

A fuel consumption of 16 km/l has a z-score of (16-15)/4=0.24. If I look this up in a table of the cumulative standard normal distribution I find:

P(z<0.25)=0.5987,

so:

P(z>0.25)=1-0.5987=0.4013,

Which is the probability P(X>16).

RonL

 

[Guest]

Hi all,

I was wondering if anybody could help me with the following questions

Suppose gas consumption for a particular type of car follows a normal distribution with mean 15 kilometres per litre and a standard deviation 4 kilometres per litre. per litre is closest to ?

b) Four hundred (400) cars of this type are selected randomly. The probability that at least 220 of them will have gas consumption over 15 kilometres per litre is closest to ?

This is what I have done:

b) smaple mean= 200
sample standard deviation = 10

Thus, P(x is greater than or equal to 220)= (219.5-200)/10
= 1.95

And then would I find the corresponding z-score value?

The number of cars in a sample of 400 with fuel consumption over 15km/l has a binomial distribution with p=0.5, N=400. Here we are working with large numbers so the normal approximation to the binomial is what we are expected to use.

220 or more in the normal approximation is 219.5, and the mean is 200, and standard deviation is 10 as you have. Thus z=(219.5-200)/10=1.95, looking this up in a table of the cumulative standard normal distribution tells us that:

P(z<1.95)=0.9744,

so:

P(z>1.95)=1-0.9744=0.0256.

So:

P(x is greater than or equal to 220)~=0.0256

RonL