A type of golf ball is tested by dropping it onto a hard surface from a height of 1 meter. The height it bounces is known to be
normally distributed. A sample of 25 balls is tested and the bounce heights are given below. Use Excel to find a 95% confidence interval for the mean bounce height of the golf ball. Round your answers to two decimal places and use increasing order.
Height
81.4
80.8
84.4
85.6
82.9
76.0
80.0
83.2
80.8
79.6
82.9
83.4
82.2
86.0
76.2
84.8
82.0
76.3
77.0
75.4
82.0
79.8
80.4
86.9
82.1
Height
81.4
80.8
84.4
85.6
82.9
76.0
80.0
83.2
80.8
79.6
82.9
83.4
82.2
86.0
76.2
84.8
82.0
76.3
77.0
75.4
82.0
79.8
80.4
86.9
82.1
1 answer:
Answer:
79.95, 82.62
Step-by-step explanation:
using excel to find a 95% confidence interval for the mean bounce height of the golf ball
Heights given are :
81.4
80.8
84.4
85.6
82.9
76.0
80.0
83.2
80.8
79.6
82.9
83.4
82.2
86.0
76.2
84.8
82.0
76.3
77.0
75.4
82.0
79.8
80.4
86.9
82.1
The statistical out put of the problem after solving with excel is attached below
therefore the 95% confidence interval from the attached solution will be ( 79.95, 82.62 )
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Pictures always help; see the one I've attached for reference.
Call the first vector (from HQ to the supply drop) u and the second vector (from supply drop to medics) v. We then want to find w, the vector from the medics to HQ, which corresponds to the vector -(u + v). (This is because u + v is the vector pointing from HQ to the medics; we're talking about the one in the opposite direction.)
Write the vectors in horizontal/vertical component form:
u = (125 km) (cos 235º x + sin 235º y) = (-71.70 x - 102.39 y) km
v = (75 km) (cos 110º x + sin 110º y) = (-25.65 x + 70.48 y) km
Why these angles?
- "55 degrees south of west" is 235º; due west is 180º from the positive horizontal axis, and you add 55º to this
- "20 degrees west of north" is 110º; due north is 90º, so add 20º to this
Add the vectors:
u + v = (-97.35 x - 31.92 y) km
Multiply by -1 to get the vector w:
w = -(u + v) = (97.35 x + 31.92 y) km
The distance covered by this vector is equal to its magnitude:
||w|| = √((97.35 km)^2 + (31.92 km)^2) = 102.45 km
The direction <em>θ</em> is given by
tan<em>θ</em> = (31.92 km)/(97.35 km) ==> <em>θ</em> = 18.15º
