Given: Security Service Company:
1.4 1.8 1.6 1.7 1.5 1.5 1.7 1.6 1.5 1.6
Mean: 1.59
Standard Deviation: 0.014333
Other companies: 1.8 1.9 1.6 1.7 1.6 1.8 1.7 1.5 1.8 1.7
Mean: 1.71
Standard deviation: 0.014333
The coefficient of variation for security Service Company:
CV = (Standard Deviation/Mean) * 100.
= (0.14333/1.59) * 100
= 9.01%
The coefficient of variation for other companies:
CV = (Standard Deviation/Mean) * 100.
= (0.014333 / 1.71) * 100
= 8.38%
the limited data listed here show evidence of stealing by the security service company's employees because there is a significant difference in the variation.
Answer:
The answer to the question is;
The probability that the resulting sample mean of nicotine content will be less than 0.89 is 0.1587 or 15.87 %.
Step-by-step explanation:
The mean of the distribution = 0.9 mg
The standard deviation of the sample = 0.1 mg
The size of the sample = 100
The mean of he sample = 0.89
The z score for sample mean is given by
where
X = Mean of the sample
μ = Mean of the population
σ = Standard deviation of the population
Therefore Z =
= -1
From the standard probabilities table we have the probability for a z value of -1.0 = 0.1587
Therefore the probability that the resulting sample mean will be less than 0.89 = 0.1587 That is the probability that the mean is will be less than 0.89 is 15.87 % probability.
(4,-2)(5,0)
slope(m) = (0 - (-2) / (5 - 4) = 2/1 = 2
there is going to be 2 possible answers...
y - y1 = m(x - x1)
now we sub
using points (4,-2)..... y -(-2) = 2(x - 4) =
y + 2 = 2(x - 4) <==
using points (5,0)....y - 0 = 2(x - 5) <==
P=2*(l+w)
p=2l+2w
2w=p - 2l
w= (p-2l)/2
Answer:
true
Step-by-step explanation: