Answer:
a) P ( X ≤ 49 ) = 0.9149
b) P ( X ≥ 48 ) = 0.2119
c) 0.9876
Explanation:
Solution:-
- Lets define a random variable X: the maximum speed of a moped.
- The random variable follows normal distribution with the following parameters:
X ~ Norm ( u , σ^2 )
Where,
u = Mean = 46.6 km/h
σ = Standard deviation 1.75 km/h
Hence,
X ~ Norm ( 46.6 , 1.75^2 ).
a) The probability that the maximum speed of mopeds is atmost 49 km/h?
- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:
- Now use the standard normal tables to determine the required probability of:
P ( Z < 1.37142 ) = 0.9149
Hence,
P ( X ≤ 49 ) = 0.9149
b) The probability that the maximum speed of mopeds is at-least 48 km/h?
- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:
- Now use the standard normal tables to determine the required probability of:
P ( Z ≥ 0.8 ) = 0.2119
Hence,
P ( X ≥ 48 ) = 0.2119
c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.
- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.
- We don't need to evaluate the test statistics as we are already given the standard error about mean.
- Using, the standard normal Z-score: The required probability is:
P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1
= 2*0.9938 - 1
= 0.9876
Answer:
C) 16.3 ml
Explanation:
Density is equal to the ratio between the mass of an object and its volume:
where
m is the mass
V is the volume
In our problem, we know:
- density of aluminium:
- mass of the aluminium foil:
So we can re-arrange the equation above and use these data to find the volume of the piece of aluminium foil:
a) Electric field inside the paint layer: zero
b) Electric field just outside the paint layer:
c) Electric field 8.00 cm outside the paint layer:
Explanation:
a)
We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:
where
E is the electric field
dS is the element of surface
q is the charge within the gaussian surface
is the vacuum permittivity
Here we want to find the electric field just inside the paint layer, so we take a sphere of radius as Gaussian surface, where
R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)
By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore
So we have
which means that the electric field inside the paint layer is zero.
b)
Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius
The area of the surface is
And since the electric field is perpendicular to the surface at any point, Gauss Law becomes
The charge included within the sphere in this case is the charge on the paint layer, therefore
So, the electric field is:
where the negative sign means the direction of the field is inward, since the charge is negative.
c)
Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.
Therefore, we have to take a Gaussian sphere of radius:
Gauss theorem this time becomes
And the charge included within the sphere is again the charge on the paint layer,
Therefore, the electric field is
Learn more about electric field:
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