A sample of 8 students was asked how often they used campus dining facilities during the past month. The responses were as follo
1 answer:
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Answer:
87.29%
Step-by-step explanation:
Given: Mean= 68.8 inches
Standard deviation= 2.8 inches
Now, finding the percent of the men in the sample were greater than 72 inches.
We know, z-score=
z-score=
⇒ z-score=
∴ z-score= 1.14
Next, using normal distribution table to find percentage.
Coverting 0.8729 into percentage=
We get the percentage as 87.29%
Hence, 87.29% of the men in the sample were greater than 72 inches.