Answer to Question #116704 in Statistics and Probability for Wale

Question

A golfer recorded the following scores in 25 recent games:
88, 89, 102, 93, 99, 95, 94, 96, 84, 91, 95, 86, 95, 92, 102, 84, 104, 108, 92, 93, 92, 97, 96, 95, 82
a) Determine the mean and standard deviation
b) Assuming a normally distributed set of scores, what is the probability that the golfer will qualify for the club championship, which requires him to get a score under 83 on his next game?

Answer 1

$82,84,84,86,88,89,91,92,92,92,93,93,94,$

$95,95,95,95,96,96,97,99,102,102,104,108$

a)

$$mean=\bar{X}={1\over 25}(82+84+84+86+88+89+91+$$ $$92+92+92+93+93+94+95+95+95+95$$ $$+96+96+97+99+102+102+104+108)=93.76$$

$$\displaystyle\sum_{i=1}^{25}(X_i-\bar{X})^2=(82-93.76)^2+2(84-93.76)^2+$$ $$+(86-93.76)^2+(88-93.76)^2+(89-93.76)^2$$

$$+(91-93.76)^2+3(92-93.76)^2+2(93-93.76)^2+$$

$$+(94-93.76)^2+4(95-93.76)^2+2(96-93.76)^2+$$

$$+(97-93.76)^2+(99-93.76)^2+2(102-93.76)^2+$$

$$+(104-93.76)^2+(108-93.76)^2=960.56$$

$$s^2={\displaystyle\sum_{i=1}^{25}(X_i-\bar{X})^2\over n-1}={960.56\over 25-1}={120.07\over 3}\approx40.023333$$

$$s=\sqrt{s^2}=\sqrt{{120.07\over 3}}\approx6.3264$$

b)

$X\sim N(\mu, \sigma^2).$ Then

$$Z={X-\mu\over \sigma}\sim N(0, 1)$$

$\mu=93.76, \sigma=6.3264$

$$P(X>83)=1-P(X\leq 83)=1-P(Z\leq {83-93.76\over 6.3264})\approx$$

$$\approx1-P(Z\leq -1.7008)\approx0.9554$$