Pregnancy length (in days) is a normally distributed random variable with a mean of 266 days and a standard deviation of 16 days
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Answer:
The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).
Step-by-step explanation:
We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of .
So it is z with a pvalue of , so
Now, find M as such
In which is the standard deviation of the population and n is the size of the sample.
The lower end of the interval is the mean subtracted by M. So it is 49.57 - 12.17 = $37.40.
The upper end of the interval is the mean added to M. So it is 49.57 + 12.17 = $61.74.
The 98% confidence interval for the mean purchases of all customers is ($37.40, $61.74).
<em>Greetings from Brasil...</em>
In a trigonometric function
F(X) = ±UD ± A.COS(Px + LR)
UD - move the graph to Up or Down (+ = up | - = down)
A - amplitude
P - period (period = 2π/P)
LR - move the graph to Left or Right (+ = left | - = right)
So:
A) F(X) = COS(X + 1)
standard cosine graph with 1 unit shift to the left
B) F(X) = COS(X) - 1 = -1 + COS(X)
standard cosine graph with 1 unit down
C) F(X) = COS(X - 1)
standard cosine graph with shift 1 unit to the right
D) F(X) = SEN(X - 1)
standard Sine graph with shift 1 unit to the right
Observing the graph we notice the sine function shifted 1 unit to the right, then
<h3>Option D</h3><em>(cosine star the curve in X and Y = zero. sine start the curve in Y = 1)</em>
