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The measures of the diameters of the heads of 150 screws is given in
the following table. If the mean diameter of the heads of the screws is
51.2 mm, find the values of p and q:
Diameter 32 - 36 37-41 42-46 47-51 52 - 56 57 - 61 62 - 66
fin mm)
No. of screws 15 17 p 25
20 30
Answers
p = 32 & q = 11
Step-by-step explanation:
Diameter Mean Diameter Frequency Mean*frequency
32 - 36 34 15 510
37 - 41 39 17 663
42 - 46 54 p 54p
47 - 51 49 25 1225
52 - 56 54 20 1080
57 - 66 59 30 1770
62 - 56 64 q 64q
107+p + q 5248+54p + 64q
107 + p + q = 150
=> p + q = 43 Eq1
Mean = (5248+54p + 64q)/150
=> 51.2 = (5248+54p + 64q)/150
=> 54p + 64q = 2432
=> 27p + 32q = 1216 Eq2
Eq2 - 27*eq1
=> 5q = 55
=> q = 11
q = 11
p = 32
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Answer:
therefore p is equal to 12 and q is equal to 35
