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The number of golf balls ordered by customers of a pro shop has the following probability distribution.
x p(x)
3 0.14
6 0.03
9 0.36
12 0.37
15 0.10
a. What are the requirements for a probability distribution table?
A. x is a random variable
B. The sum of P(x) must equal 1
C. Each probability p(x) must be between 0 and 1 inclusive
D. All the above
b. Find the mean for the random variable x. (round to a whole number)
c. Find the standard deviation of the random variable x. (round to a whole number)
d. Based on the table results would be significanlty high to place an order of 15 golf balls?
A. No, 15 is not a significantly high number of golf balls, because the probability of 15 is greater than 0.05
B. Yes, 15 is a significantly high number of golf balls, because the probability of 15 is less than 0.05
C. No, 15 is not a significantly high number golf balls, because the probability of 15 is less than 0.05
D. Yes, 15 is a significantly high number of golf balls, because the probability of 15 is greater than 0.05
E. No, 15 is not a significantly high number of golf balls, it is a low number of golf balls

Sagot :

MrRoyal

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

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