225k views
5 votes
A golfer keeps track of his score for playing nine holes of golf​ (half a normal golf​ round). His mean score is 8080 with a standard deviation of 1313. Assuming that the second 9 has the same mean and standard​ deviation, what are the mean and standard deviation of his total score if he plays a full 18​ holes?

User Brubs
by
6.7k points

1 Answer

6 votes

Answer:

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and
11\sqrt2 respectively.

Explanation:

Given that,

For the first 9 holes X:

E(X) = 80

SD(X)=13

For the second 9 holes Y:

E(Y) = 80

SD(Y)=13

For the sum W=X+Y, the following properties holds for means , variance and standard deviation :

E(W)=E(X)+E(Y)

and

V(W)=V(X)+V(Y)

⇒SD²(W)=SD²(X)+SD²(Y) [ Variance = (standard deviation)²]


\Rightarrow SD(W)=√(SD^2(X)+SD^2(Y))

∴E(W)=E(X)+E(Y) = 80 +80=160

and


SD(W)=√(SD^2(X)+SD^2(Y))


=√(11^2+11^2)


=√(2.11^2)


=11\sqrt2

Therefore the mean and standard deviation of his total score if he plays a full 18 holes are 160 and
11\sqrt2 respectively.

User Clime
by
6.4k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.