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arsen [322]
3 years ago
11

Suppose your manager indicates that for a normally distributed data set you are analyzing, your company wants data points betwee

n z = − 1.6 z=-1.6 and z = 1.6 z=1.6 standard deviations of the mean (or within 1.6 standard deviations of the mean). What percent of the data points will fall in that range?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

89.04% of the data points will fall in the given range of z = − 1.6 and z = 1.6      

Step-by-step explanation:

We are given a normally distributed data.

We have to find the percentage of data that lies within the range  z = − 1.6 and z= 1.6

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(-1.6 \leq z \leq 1.6)\\= P(z \leq 1.6) - P(z \leq -1.6)\\\text{Calculating the value from standard normal table}\\= 0.9452 - 0.0548 = 0.8904= 89.04\%

89.04% of the data points will fall in the given range of z = − 1.6 and z= 1.6

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The function of the table is quadratic in nature and is given as follows:   y = 2x² -1

<h3>What is a quadratic function?</h3>

A quadratic function is one of the following: f(x) = ax² + bx + c, where a, b, and c are positive integers and a, b, and are not equal to zero.

Given the nature of a quadratic function (y =  ax² + bx + c) we pick three pairs of (x, y) from the table as follows:

A) (0, -1)

B) (1, 1)

C) (2, 7)

Then we say:

A) (0, -1) →   -1 = a(0)²  + b(0) + c =

-1 = c ..................................1

B) (1, 1) →   1 = a(1)² + b(1) + c =

1 = a + b + c.......................2

C) (2, 7) → 7 = a(2)² + b(2) + c

7 = a (4) + 2b + c

7 = 4a + 2b + c .................3

Using elimination and substitution, let us substitute equation 1 into equation 2;

that is

1 = a + b + (-1)

1 = a + b -1
1 + 1 = a + b

2 = a + b................................4

From the above we can also say,

b = 2 - a...................................5

substitute 5 into 3 we have

7 = 4a + 2 (2-a) + (-1)

7 = 4a + 4 - 2a  - 1

7 = 4a - 2a + 4 - 1

7 = 2a + 3

7- 3 = 2a

4 = 2a

a = 4/2

a = 2

taking [a = 2] and [c = -1], subsitute these into equation 2

1 = a + b + c

1 = 2 + b + (-1)

1 = 2 + b - 1

1 = 2 - 1 + b

1 = 1 + b

b = 1-1

b= 0

Hence,

a = 2

b = 0

c = -1

Hence:

y = 2 * x² + (0) x + (-1)

y = 2x² -1

Hence the equation for the above table is y = 2x² -1

Learn more about Quadratic Functions:
brainly.com/question/1214333

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