For a standard normal distribution find the approximate value of p(z<0.42)
2 answers:
Answer:
p(z<0.42) = 0.6628
Step-by-step explanation:
A normal distribution with a mean of 0 and a standard deviation of 1 is called a standard normal distribution. That is to say:
μ = 0
σ² = 1
Using a calculator, we find that:
p(z<0.42) = 0.6628 (See picture attached)
Answer:
0.66276.
Step-by-step explanation:
We are asked to find the approximate value of p(z<0.42) for a standard normal distribution.
Our given expression means the probability of getting a z-score less than 0.42.
We need to find the probability of getting the area corresponding to a z-score less than 0.42 under normal distribution curve.
We will normal distribution table to solve our given problem.
Therefore, the approximate value of our given expression would be 0.66276.
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Answer:
72 sq. units
Step-by-step explanation:
Find the diagram attached
The rhombus is the area of the two triangles and the rectangle
Area of triangle = 1/2 * base * height
Area of triangle = 1/2 * y * h
Area of triangle = 1/2 * 3 * 8
Area of triangle = 3 * 4
Area of triangle = 12 sq.units
Area of 2 triangle = 2(12) = 24 sq. units
Area of the rectangle = length * width]
Area of the rectangle = h (x-y)
Area of the rectangle = 8(9-3)
Area of the rectangle = 8 * 6
Area of the rectangle = 48 sq. units
Area of the rhombus = 24 + 48
Area of the rhombus = 72 sq. units
