Question

1. the golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. let x = golf score for a random school team, then fill in the blanks: then x ~ n (68, 3). a. find the probability that a randomly selected golfer scored at most 65. p(x≤65)=normalcdf(-e99,65,68,3) b. find the proportion (or probability) that a randomly selected golfer scored more than 80. p(x > 80)=normalcdf(80,e99,68,3) c. find the proportion (or probability) that a golfer scored between 65 and 70. p(65 < x < 70)=normalcdf(65,70,68,3) d. find the 30th percentile, and interpret it in a complete sentence. invnorm(.3,68,3) interpret: 30% of the golf scores scored 66.43 e. find the minimum of the third quartile for this distribution. q3 = invnorm(.75,68,3) f. the middle 30% of the golf scores are between what two values? y1 = invnorm(.35,68,3); y2 = invnorm(.65,68,3) g. the highest 10% golf scores are above what value? invnorm(,68,3)

Explanation:

Step1: Standardize the value for part a

We use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 68$, $\sigma=3$ and $x = 65$. So $z=\frac{65 - 68}{3}=\frac{- 3}{3}=-1$. Then $P(X\leq65)=P(Z\leq - 1)$. Looking up in the standard - normal table, $P(Z\leq - 1)=0.1587$.

Step2: Standardize the value for part b

For $x = 80$, $z=\frac{80 - 68}{3}=\frac{12}{3}=4$. Then $P(X>80)=1 - P(X\leq80)=1 - P(Z\leq4)$. Since $P(Z\leq4)\approx1$, $P(X>80)\approx1 - 1=0$.

Step3: Standardize the values for part c

For $x_1 = 65$, $z_1=\frac{65 - 68}{3}=-1$ and for $x_2 = 70$, $z_2=\frac{70 - 68}{3}=\frac{2}{3}\approx0.67$. Then $P(65

Step4: Find the z - score for the 30th percentile in part d

We look up the z - score $z$ in the standard - normal table such that $P(Z\leq z)=0.3$. The z - score is approximately $z=-0.52$. Then, using the formula $x=\mu+z\sigma$, we have $x = 68+( - 0.52)\times3=68 - 1.56 = 66.44\approx66.43$. Interpretation: 30% of the golf scores are 66.43 or less.

Step5: Find the z - score for the third quartile in part e

The third quartile corresponds to the 75th percentile. Looking up the z - score in the standard - normal table for $P(Z\leq z)=0.75$, we get $z\approx0.67$. Then $x=\mu+z\sigma=68 + 0.67\times3=68+2.01 = 70.01\approx70.02$.

Step6: Find the z - scores for part f

For the lower bound of the middle 30%, we want the area to the left to be $0.35$. The z - score $z_1\approx - 0.39$. For the upper bound, the area to the left is $0.65$, and the z - score $z_2\approx0.39$. Then $x_1=\mu+z_1\sigma=68+( - 0.39)\times3=68 - 1.17 = 66.83\approx66.84$ and $x_2=\mu+z_2\sigma=68+0.39\times3=68 + 1.17 = 69.17\approx69.16$.

Step7: Find the z - score for part g

The area to the left for the value we want is $0.9$. Looking up the z - score in the standard - normal table, $z\approx1.28$. Then $x=\mu+z\sigma=68+1.28\times3=68 + 3.84 = 71.84$.

Answer:

a. $0.1587$
b. $0$
c. $0.59$
d. $66.43$; 30% of the golf scores are 66.43 or less.
e. $70.02$
f. $66.84$ (Low) and $69.16$ (High)
g. $71.84$