Answer:
(a) The probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b) The probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Step-by-step explanation:
Let the random variable <em>X</em> follow a Normal distribution with parameters <em>μ</em> = 155.4 and <em>σ</em> = 49.5.
(a)
Compute the probability that a single randomly selected value lies between 158.6 and 159.2 as follows:

*Use a standard normal table.
Thus, the probability that a single randomly selected value lies between 158.6 and 159.2 is 0.004.
(b)
A sample of <em>n</em> = 246 is selected.
Compute the probability that a sample mean is between 158.6 and 159.2 as follows:

*Use a standard normal table.
Thus, the probability that a sample mean is between 158.6 and 159.2 is 0.0411.
Well try working on the problem first
In intercept form, the plane that has these intercepts is ...
... x/(x-intercept) + y/(y-intercept) + z/(z-intercept) = 1
... x/1 + y/(-1) + z/2 = 1
... 2x -2y +z = 2 . . . . . in standard form
The direction vector of the line
L: x=1+t, y=4t, z=2-3t
is <1,4,-3>
which is also the required normal vector of the plane.
Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1
This would be 11.83 I hope you pass your test/quiz :)