In 2002, the Centers for Disease Control and Prevention (CDC) reported that 8% of women married for the first time by their 18th
1 answer:
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Answer:
(a) 0.0833 or 8.33%
(b) 0.40 or 40%
Step-by-step explanation:
Parts line one (n1) = 1,000 parts
Defects line one (d1) = 100 parts
Parts line two (n2) = 2,000 parts
Defects line two (d2) = 150 parts
Total number of parts (n) = 3,000 parts
a. Probability of a randomly selected part being defective:
The probability is 0.0833 or 8.33%
b. Probability of a part being produced by line one, given that it is defective:
The probability is 0.40 or 40%.
Answer:
a) P-value = 0.0968
b) P-value = 0.2207
c) P-value = 0.0239
d) P-value = 0.0040
e) P-value = 0.5636
Step-by-step explanation:
As the hypothesis are defined with a ">" sign, instead of an "≠", the test is right-tailed.
For this type of test, the P-value is defined as:
being z* the value for each test statistic.
The probability P is calculated from the standard normal distribution.
Then, we can calculate for each case:
(a) 1.30
(b) 0.77
(c) 1.98
(d) 2.65
(e) −0.16
Answer:
The median and mode of the students grade is 70.
Most of the students scored between 40 and 100.
Step-by-step explanation:
From the provided information it can be seen that the mean of the distribution is, <em>μ</em> = 70 and the standard deviation is, <em>σ</em> = 15.
For a Normal distributed data the mean, median and mode are the same.
So, the median and mode of the students grade is 70.
The standard deviation of the data represents the spread of the observation, i.e. how dispersed the values are along the curve.
In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68.27%, 95.45% and 99.73% of the values of a Normally distributed data lie within one, two and three standard deviations of the mean, respectively.
Assuming that maximum marks of the exam is 100, it can be said that most of the students scored between 40 and 100.