All categories

2 years ago

(c) Use a calculator to verify that Σ(x) = 64, Σ(x2) = 1118, Σ(y) = 628, Σ(y2) = 87,982, and Σ(x y) = 9,627.

Mathematics

1 answer:

amid [387]2 years ago

7 0

The correlation based on the computation is 0.97.

<h3>How to compute the value?</h3>

It should be noted that the question is that we should simply calculate the correlation which is r.

The correlation will be:

= Σ(x y) / (Σ(x²) × Σ(y²)

= 9627/✓(1118 × 87982)

= 9627 ✓98363876

= 9627/9917.86

= 0.97

Learn more about computations on:

brainly.com/question/4658834

#SPJ1

You might be interested in

A group of 200 people who exercise regularly reported the following exercise habits :

xxMikexx [17]

Answer:

9/20

Step-by-step explanation:

65 people swim or run

BUT

5 people just run

60 people swim or run

30 people swim and run

90/200 =

9/20

7 0

2 years ago

FIND THE AREA OF THE SHADED SECTION IN THE CIRCLE THE SHADED SECTION IS 35 DEGRESS AND THE RADIUS IS 5 INCHES. USE THE FORMULA:

Natali [406]

Answer:

Area of sector of circle = 7.6302 inches (Approx.)

Step-by-step explanation:

Given:

Angle of section = 35°

Radius of circle = 5 inch

Find:

Area of shaded section

Computation:

Area of shaded section = Area of sector of circle

Area of sector of circle = [θ/360][πr²]

Area of sector of circle = [35/360][(22/7)(5)²]

Area of sector of circle = [0.0972][(3.14)(25)]

Area of sector of circle = [0.0972][78.5]

Area of sector of circle = 7.6302 inches (Approx.)

6 0

3 years ago

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

yKpoI14uk [10]

Answer:

a) 0.5588

b) 0.9984

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24

Standard Deviation, σ = 6.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

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

a) P(score between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{6.4} \leq z \leq \displaystyle\frac{30-24}{6.4}) = P(-0.62 \leq z \leq 0.94)\\\\= P(z \leq 0.94) - P(z < -0.62)\\= 0.8264 - 0.2676 = 0.5588 = 55.88\%$

$P(20 \leq x \leq 30) = 55.88\%$

b) Sampling distribution

Sample size, n = 22

The sample will follow a normal distribution with mean 24 and standard deviation,

$s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{6.4}{\sqrt{22}} =1.36$

c) P(mean score of sample is between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{1.36} \leq z \leq \displaystyle\frac{30-24}{1.36}) = P(-2.94 \leq z \leq 4.41)\\\\= P(z \leq 4.41) - P(z < -2.94)\\= 1 - 0.0016 = 0.9984 = 99.84\%$