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06 Jul 2010, 12:49
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Hi,
Can someone advise me if my calculation for the below question is correct.
Q1. An HR Manager is concerned about the absenteeism of hourly employees, especially on Wednesday in the summer months. He has collected data for Wednesday in May over the past 6 years as follows:
No. absent No. absent No. absent
103 96 114
91 84 69
100 98 89
87 97 93
65 104 88
104 112 102
119 88 109
103 96 93
Find the mean number of employees absent as well as the standard deviation. If absenteeism on Wednesday in May is felt to be normally distributed, what is the probability there would be more than 112 people absent on a Wednesday in May?
My calculation:
A. Mean = 96
B. Standard Deviation = 12.41 or 12
C. Probability of more than 112 people absent on Wednesday in month of May
where:
mean = 96
std. dev. = 12
X = 112
Z score computation:
z= X - mean / std.dev
z = 112 - 96 / 12
z = 1.33 = 0.4082
On Z distribution table, the value of 1.33 is 0.4082
Since X is greater than the mean value
P(X) = 0.5 + Z = 0.5 + 0.4082 = 0.91 or 91%
Appreciate for a positive feedback.
Thanks.
Can someone advise me if my calculation for the below question is correct.
Q1. An HR Manager is concerned about the absenteeism of hourly employees, especially on Wednesday in the summer months. He has collected data for Wednesday in May over the past 6 years as follows:
No. absent No. absent No. absent
103 96 114
91 84 69
100 98 89
87 97 93
65 104 88
104 112 102
119 88 109
103 96 93
Find the mean number of employees absent as well as the standard deviation. If absenteeism on Wednesday in May is felt to be normally distributed, what is the probability there would be more than 112 people absent on a Wednesday in May?
My calculation:
A. Mean = 96
B. Standard Deviation = 12.41 or 12
C. Probability of more than 112 people absent on Wednesday in month of May
where:
mean = 96
std. dev. = 12
X = 112
Z score computation:
z= X - mean / std.dev
z = 112 - 96 / 12
z = 1.33 = 0.4082
On Z distribution table, the value of 1.33 is 0.4082
Since X is greater than the mean value
P(X) = 0.5 + Z = 0.5 + 0.4082 = 0.91 or 91%
Appreciate for a positive feedback.
Thanks.
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06 Jul 2010, 14:42
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First of all it's not a GMAT question.
Your solution seems to be right except one thing: the question asks "...would be more than 112 people...". So, it's rather 100 - 91 = 9%
Your solution seems to be right except one thing: the question asks "...would be more than 112 people...". So, it's rather 100 - 91 = 9%
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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