3. Supremely succulent Scottish super spy Sal Sperry Stake estimates distances f

Question

3. Supremely succulent Scottish super spy Sal Sperry Stake estimates distances for both gunshot accuracy and her golf game. Her estimation errors follow a Normal distribution with mean =-0.2m and standard deviation = 0.5m. a. How should you interpret that Stake's mean estimation error is -0.2m? b. What is the probability that one of Stake's estimates is: i. more than a meter too short? ii. off by no more than 0.5m? (5,5) (10) (10) (10,10) (53) c. Over 36 holes of golf, what are the probabilities that: i. Stake overestimates distances on at least 10 holes? Explain! i. Stake's mean estimation error is negative? Dittet ii. Stake's mean estimation error is within 0.1m of correct?

Explanation / Answer

QUestion.

Error mean = -0.2 m

Stanard deviation = 0.5 m

(a) Here the estimation errror mean is -0.2 m and a normal distriubtion is there. So, We can say that the mean estimation error is -0.2m.

(b) Here we have to find that the probability that the estimates are more than a meter too short.

Pr(X < -1) = NORM(X < -1 ; -0.2 ; 0.5)

Z = (-1 + 0.2)/ 0.5 = -1.6

so Pr(X < -1) = Pr (Z < -1.6) = 0.0548

(ii) Off by no more than 0.5m

Pr(X > 0.5 m) = NORM(X> 0.5 ; -0.2 ; 0.5)

Z = (0.5 + 0.2)/ 0.5 = 1.4

Pr(X > 0.5 m) = 1 - Pr(Z > 1.4) = 1 - 0.9192 = 0.0808

(c) Here sample size n = 36

(i) Stake overestimates on at least 10 holes. That means the error must be greater than 0.

So, Here Pr(X > 0) = NORMAL (X > 0 ; -0.2 ; 0.5)

Z = ( 0 + 0.2)/ 0.5 = 0.4

Pr(X > 0) = 1 - Pr( Z < 0) = 1 - 0.6554 = 0.3446

so now it has become binomial distribution where we know that n = 36 and p = 0.3446 and Y is the random variable " Number of overestimated distance "/

Pr(Y >=10) = BIN(Y >=10; 36; 0.3446) = 1 - BIN(Y < 10 ; 36; 0.3446)

from Binomial table

Pr(Y >=10) = BIN(Y >=10; 36; 0.3446) = 1 - BIN(Y < 10 ; 36; 0.3446) = 1- 0.1538 = 0.8462

(ii) Mean estiamation error is nagative.

That mean if sample mean is x , then it shall be negative.

Pr(x< 0)

Here estimated mean of sample mean = -0.2

standard error of sample mean = / sqrt(n) = 0.5/6 = 0.0833

So,

Pr(x < 0 ; -0.2 ; 0.08333)  

Z = ( 0 + 0.2)/ 0.08333 = 2.4

Pr(x < 0 ; -0.2 ; 0.08333) = Pr(Z < 2.4) = 0.9918

(iii) Here mean estimation error is within 0.1 m of correct.

THat means, sample mean will be in between [-0.1, 0.1]

Pr( -0.1 < x < 0.1) = Pr(-0.1 < x < 0.1 ; -0.2; 0.08333) = Pr( x < 0.1 ; -0.2; 0.08333) - Pr(-0.1 < x ; -0.2; 0.08333)

Z2 = (0.1 + 0.2)/ 0.08333 = 3.6 ; Z1 = (-0.1 + 0.2)/ 0.0833 = 1.2

Pr( -0.1 < x < 0.1) = Pr(Z < 3.6) - Pr(Z < 1.2) = 0.99984 - 0.8849 = 0.1149

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