Answer:
(a) D. All the above
(b) [tex]p = 10[/tex] --- Mean
(c) [tex]s = 3[/tex] --- Standard Deviation
(d) A. No, 15 is not a significantly high number of golf balls, because the probability of 15 is greater than 0.05
Step-by-step explanation:
Solving (a):
Option (d) answers the question because (a), (b) and (c) are all true
For (a): x must be a random variable
For (b): [tex]\sum p(x) = 1[/tex]
For (c): [tex]0 \le p(x) \le 1[/tex]
Solving (b): The mean
Mean (p) is calculated as:
[tex]p = \sum xp(x)[/tex]
So, we have:
[tex]p = 3 * 0.14 + 6 * 0.03 + 9 * 0.36 + 12 * 0.37 + 15 * 0.10[/tex]
[tex]p = 9.78[/tex]
[tex]p = 10[/tex]
Solving (c): The standard deviation
This is calculated as:
[tex]s = \sqrt{\sum[ x*p(x) *(1-p(x))]}[/tex]
So, we have:
[tex]s = \sqrt{3 * 0.14 *(1 - 0.14) + 6 * 0.03 * (1 - 0.03) +..............+ 15 * 0.10 * (1 - 0.10)}[/tex][tex]s = \sqrt{6.7566}[/tex]
[tex]s = 2.599[/tex]
[tex]s = 3[/tex]
Solving (d): Is P(x = 15) significantly high?
From the question:
[tex]P(x = 15) = 0.10[/tex]
In probability:
[tex]0 \le x \le 0.15[/tex] is considered to be low (i.e. not high) because it is almost not sure to occur
Hence, (a) is true
