An autopilot attempts to maintain an airplane's altitude at a constant 18000 feet. because of random variations, the actual altitude x is a normal random variable with mean 18000 feet and variance=2500 feet2 . (a) what is the probability that the altitude differs from the assigned 18000 value by more than 100 feet? (b) suppose that by mechanically improving the autopilot, the variance of the airplane's altitude can actually be reduced. find the value of the variance such that the probability that the altitude differs from the assigned 18000 value by more than 100 feet is at most 0.01

(a) A suitable probability app tells you the probability of being less than 17,900 or more than 18,100 is about 4.55%.

(b) We want to find σ² such that 2P((18100-18000)/σ) ≤ 0.01. We use a factor of 2 because we want to count deviations both above and below 100 feet from the mean altitude. A suitable probability app can tell you the z-score such that P(Z)≤0.005, so we need to compute σ² = (100/Z)² from that result. Doing so, we get σ² ≈ 1507.

the required function is .

step-by-step explanation:

from the given table it is noticed that the graph passing through (-3,7), (0,4), (3,1) and (8,-4).

it is a linear function because the value of y is decrease by one as the value of x increase by 1.

the slope of the function is

the slope of the function is -1 and the y-intercept is 4.

the slope intercept form is

where, m is slope and b is y-intercept.

therefore the required function is . the graph of this function is shown below.

3x + 4y -25 =0 represents the amount of change sianna should recieve