a golf course is trying to determine par for a particular hole. calculate the value they should put on the sign where people tee
off so they know what par is for this hole.
1 answer:
The value they should put on the sign where people tee off so they know
what par is for the hole is six (6)
The reasoning through which the correct answer is arrived at is presented here given table of values which is presented as follows;
The mean number of golf strokes for the hole is ∑(f·x)/∑f
The calculation for f × x on the table includes;
∑(f·x) = (2×16 + 3×21 + 4×23 + 5×20 + 6×54 + 7×40 + 8×23 + 9×16) = 1219
∑f = 16 + 21 + 23 + 20 + 54 + 40 + 23 + 16 = 213
∴ ∑(f·x)/∑f = 1,219/213 = 5.72 ≈ 6
The mean number of golf strokes expected is approximately 6 strokes
The value they should put on the sign where people tee off so they know what par is for the hole is 6
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Answer:
(a) The cumulative frequency curve for the data is attached below.
(b) (i) The inter-quartile range is 10.08.
(b) (ii) The 70th percentile class scores is 0.
(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.
Step-by-step explanation:
(a)
To make a cumulative frequency curve for the data first convert the class interval into continuous.
The cumulative frequencies are computed by summing the previous frequencies.
The cumulative frequency curve for the data is attached below.
(b)
(i)
The inter-quartile range is the difference between the third and the first quartile.
Compute the values of Q₁ and Q₃ as follows:
Q₁ is at the position:
The class interval is: 34.5 - 39.5.
The formula of first quartile is:
Here,
l = lower limit of the class consisting value 25 = 34.5
(CF) = cumulative frequency of the previous class = 24
f = frequency of the class interval = 20
h = width = 39.5 - 34.5 = 5
Then the value of first quartile is:
The value of first quartile is 34.75.
Q₃ is at the position:
The class interval is: 44.5 - 49.5.
The formula of third quartile is:
Here,
l = lower limit of the class consisting value 75 = 44.5
(CF) = cumulative frequency of the previous class = 74
f = frequency of the class interval = 15
h = width = 49.5 - 44.5 = 5
Then the value of third quartile is:
The value of third quartile is 44.83.
Then the inter-quartile range is:
Thus, the inter-quartile range is 10.08.
(ii)
The maximum upper limit of the class intervals is 69.5.
That is the maximum percentile class score is 69.5th percentile.
So, the 70th percentile class scores is 0.
(iii)
Compute the probability that a student scored at most 50 on the examination as follows:
Thus, the probability that a student scored at most 50 on the examination is 0.89.
Hope that helps. and im pretty sure its 180
