discussionwk9

Data 605 discusstion wk9 # 9.2.7

7. A surveying instrument makes an error of -2, -1, 0, 1, or 2 feet with equal probabilities when measuring the height of a 200-foot tower.

a) Find the expected value and the variance for the height obtained using this instrument once.

X = height obtained Y = error X = 200 + Y

\(E(Y) = \frac{1}{5} (-2+-1+0+1+2) = 0\) E(X) = 200 + E(Y) = 200 The expected value is 200 for height obtained.

\(V(Y) = \frac{1}{5} (-2^{2}-1^{2}+0^{2}+1^{2}+2^{2}) = \frac{10}{5} =2\) V(X) = V(Y) = 2

The variance is 2.

b) Estimate the probability that in 18 independent measurements of this tower, the average of the measurements is between 199 and 201, inclusive.

\(S_{18}=X_{1}+X_{2}+X_{3}...X_{18}\)

\(average height=\frac{S^{n}}{n}\)

E <- 200
V <- sqrt((1/18)*2)
V

## [1] 0.3333333

#P(199<=Sn*<=201)

#P(199<=Sn*)

P199 <- ((199-E)/V)
P199

## [1] -3

#P(Sn*<=201)
P201 <- ((201-E)/V)
P201

## [1] 3

(0.5-dnorm(P199))*2

## [1] 0.9911363

probability in 18 independent measures of tower is 0.99113.