Question

A bowler believes he bowls a better score during his second game than his first game 74% of the time. Assuming he is right, what is the probability that in his next 25 pairs of games, he will be correct 18 to 20 times, inclusive?
a) Determine the exact probability using the binomial distribution.
b) Determine the probability using the normal approximation and compare your answer to the answer from part a).

Answer 1

\[
\Pr(18\leq X\leq 20)
\]

Answer 2

It seems like a binomial distribution, with 25 trials.

Answer 3

\[
\Pr(X=k) = \binom{25}{k}(0.74)^k(0.16)^{25-k}
\]

Answer 4

This comes from: \[
\Pr(X=k) = \binom{n}{k}p^k(1-p)^{n-k}
\]Where \(n=25\) and \(p=0.74\).

Answer 5

Since it's a discreet probability distribution:\[ \Large
\Pr(18 \leq X \leq 20) = \sum_{k=18}^{20}\Pr(X=k)
\]

Answer 6

@hager Get it?

Answer 7

@wio ....discrete probability.........

Answer 8

???

Answer 9

@hager wio is correct regarding section a)
b)The binomial distribution can be approximated by the normal distribution in this case where:
\[mean=n \times p\]
and
\[standard\ deviation=\sqrt{np(1-p)}\]

Answer 10

thnx guys

Answer 11

@wio Note that in your first equation (1 - p) should be 0.26 and not 0.16.

Answer 12

ok thnx

Answer 13

wio i didn't get for the K

Answer 14

@hager k is the number of times the player's assumption (that the second game will be better than the first game) is correct. In this problem k=18 or 19 or 20 in the series of calculations.

Answer 15

so i have to calculate them by these three numbers

Answer 16

Yes. The first calculation is:
\[P(correct\ for\ 18\ games)=\left(\begin{matrix}25 \\ 18\end{matrix}\right)0.74^{18}0.26^{7}=you\ can\ calculate\]

Answer 17

ok but i have only one result here not three diffrent answers a) 50.33% and b) 49.58%

Answer 18

but i don't know how

Answer 19

You will get the correct answer to a) by adding together the 3 values of probability obtained for k = 18, k = 19 and k = 20.

Answer 20

ok i got it thnx so much

Answer 21

You're welcome :)