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My name is Ann [436]
3 years ago
11

What is the approximate area of the unshaded region under the standard normal curve below? Use the portion of the standard norma

l table given to help answer the questions. See photo’s attached.

Mathematics
1 answer:
Charra [1.4K]3 years ago
7 0

Concept:

First find the area of the shaded region under the standard normal curve and after it as you know total area=1 , so 1-area of shaded region= area of unshaded region.

Answer:

Area of shaded region= P(-2 \leq z \leq 1)


Now,


the symbol Ф represent the cumulative density.


first find the


Ф(1) from the above given table it is equal to 0.8413.


Now,


find the Ф(-2) .


in our table we are given the value of Ф(2)=0.9772.

so as the curve is symmetrical Ф(-2)=1-0.9772=0.0228.

P(-2 \leq z \leq 1)
= Ф(1)-Ф(-2)


= 0.8413-0.0228

= 0.8185

Now,

Area of unshaded region= 1-area of shaded region

= 1- P(-2 \leq z \leq 1)

= 1- 0.8185

= 0.1815

= 0.18

C is the correct answer.

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3 years ago
1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than
zaharov [31]

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 3, n = 8

$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)

           $=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)

           $=\left(\frac{81}{11} , \frac{-43}{11}\right)

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 7, n = 1

$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)

           $=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)

           $=\left(\frac{136}{8} , \frac{-12}{8}\right)

C(x, y) = (17, –1.5)

8 0
3 years ago
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