Solve the simultaneous equation 28x+32y=88 and 32x+28y=92
1 answer:
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Answer:
(a) 283 days
(b) 248 days
Step-by-step explanation:
The complete question is:
The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. (a) What is the minimum pregnancy length that can be in the top 11% of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths?
Solution:
The random variable <em>X</em> can be defined as the pregnancy length in days.
Then, from the provided information .
(a)
The minimum pregnancy length that can be in the top 11% of pregnancy lengths implies that:
P (X > x) = 0.11
⇒ P (Z > z) = 0.11
⇒ <em>z</em> = 1.23
Compute the value of <em>x</em> as follows:
Thus, the minimum pregnancy length that can be in the top 11% of pregnancy lengths is 283 days.
(b)
The maximum pregnancy length that can be in the bottom 5% of pregnancy lengths implies that:
P (X < x) = 0.05
⇒ P (Z < z) = 0.05
⇒ <em>z</em> = -1.645
Compute the value of <em>x</em> as follows:
Thus, the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths is 248 days.
Answer:
Step-by-step explanation:
The interquartile range is IQR is:
Where:
= third quartile
= first quartile
So the interquartile range is the difference between the highest quartile and the lowest quartile. In other words, it is the width of the box in the diagram.
So:
The R range is the difference between the highest value and the lowest. In other words:
Finally: