Use the frequency distribution to find the mean and the sample standard deviation. Does the data form a normal distribution? Remember: if the data fall within 2% of the empirical rule of 68%, 95%, and 99.7% for one, two, or three standard deviations, respectively, they form a normal distribution.
Shipment time (hours)Frequency f23.5 - 28.51928.5 - 33.541733.5 - 38.578938.5 - 43.569843.5 - 48.550448.5 - 53.542
A. The mean is 45 hours. The sample standard deviation is 5 hours. The data doesn't form a normal distribution.
B. The mean is 39 hours. The sample standard deviation is 5 hours. The data doesn't form a normal distribution.
C. The mean is 39 hours. The sample standard deviation is 5 hours. The data forms a normal distribution.
D. The mean is 35 hours. The sample standard deviation is 5 hours. The data doesn't form a normal distribution.
E. The mean is 45 hours. The sample standard deviation is 5 hours. The data forms a normal distribution.
Answer: B
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The function s = f(t) gives the position of a body moving on a coordinate line, with s in meters and t in seconds.s = 3t2 + 2t + 3, 0 ? t ? 2Find the body's displacement and average velocity for the given time interval.
A. 10 m, 14 m/sec B. 22 m, 11 m/sec C. 16 m, 8 m/sec D. 16 m, 16 m/sec
Simplify.
2
A. 2
B. 
C. 
D. 
Fill in the blank with one of the words or phrases listed below.
A(n) 
is a natural number greater than 1 that is not prime.
A. numerator B. mixed number C. denominator D. composite number
Solve.(4m + 7)2 + 3(4m + 7) - 4 = 0
A. 
, 
B. - 
, 2
C. - 
, - 
D. - 
, - 2