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For her customer feedback score, Jess wants to average at or above an

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For her customer feedback score, Jess wants to average at or above an  [#permalink]

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New post 02 May 2016, 02:13
1
1
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (01:46) correct 27% (01:39) wrong based on 108 sessions

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For her customer feedback score, Jess wants to average at or above an 8.0 rating. For the month of June, she received the following scores: 7, 8, 8, 7, 9, 10, 6, 6, 8 and 7. By what percent did she fall short of her goal?

(A) 4%
(B) 5%
(C) 8%
(D) 10%
(E) 12%
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Re: For her customer feedback score, Jess wants to average at or above an  [#permalink]

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New post 02 May 2016, 06:22
Expected minimum rating = 8

Deviation from the expected minimum rating = (-1 + 0 + 0 - 1 + 1 + 2 - 2 - 2 + 0 - 1)/10 = -4/10 = -0.4

Current rating is 0.4 less than 8.

Percentage = (0.4/8)*100 = 5%

Answer: B
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Re: For her customer feedback score, Jess wants to average at or above an  [#permalink]

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New post 10 Dec 2016, 11:22
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Re: For her customer feedback score, Jess wants to average at or above an  [#permalink]

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New post 10 Dec 2016, 12:05
Nevernevergiveup wrote:
For her customer feedback score, Jess wants to average at or above an 8.0 rating. For the month of June, she received the following scores: 7, 8, 8, 7, 9, 10, 6, 6, 8 and 7. By what percent did she fall short of her goal?

(A) 4%
(B) 5%
(C) 8%
(D) 10%
(E) 12%


Total score of June is 76

She wants total score of 80

So, She fell short of 4/80*100 = 5%

Hence, answer will be (B) 5%

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Re: For her customer feedback score, Jess wants to average at or above an  [#permalink]

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New post 13 Apr 2018, 04:44
+1 for B. The trick here is to take differences over 8 and sum them up. Now once you get the net differential , calculate the %
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Re: For her customer feedback score, Jess wants to average at or above an   [#permalink] 13 Apr 2018, 04:44
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