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andriy [413]
2 years ago
12

Becky is an experienced swimmer. The table contains data on her times in the 100-meter freestyle event for each year over the pa

st six years.
Year Time
(seconds)
1 98
2 101
3 109
4 117
5 119
6 127

Sketch a scatter plot for this data, and then determine which of these numbers is the closest to the correlation coefficient for the data set.
a.0.20
b.0.99
c.-0.20
d.-0.83
e.-1.01
Mathematics
1 answer:
Sauron [17]2 years ago
5 0
Correlation coefficient (r) = [nΣxy - (Σx)(Σy)] / [sqrt(nΣx^2 - (Σx)^2)sqrt(nΣy^2 - (Σy)^2)]
Σx = 21 => (Σx)^2 = 21^2 = 441
Σy = 671 => (Σy)^2 = 671^2 = 450,241
Σx^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
Σy^2 = 98^2 + 101^2 + 109^2 + 117^2 + 119^2 + 127^2 = 75,665
Σxy = 1(98) + 2(101) + 3(109) + 4(117) + 5(119) + 6(127) = 2,452
r = [6(2,452) - 21(671)] / [sqrt(6(91) - 441)sqrt(6(75,665) - 450,241)] = 621/sqrt(105)sqrt(3749) = 0.99

option b is the correct answer.
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A fisherman catches fish according to a Poisson process with rate lambda = 0.6 per hour. The fisherman will keep fishing for two
spayn [35]

Answer:

Step-by-step explanation:

Given that a fisherman catches fish according to a Poisson process with rate lambda = 0.6 per hour.

The fisherman will keep fishing for two hours.

Since he continues till he gets atleast one fish, we can calculate probability as follows:

(a) Find the probability that he stays for more than two hours.

= Prob (x=0) in I two hours and P(X≥1) in 3rd hour

=P(x=0)*P(X=0)*P(X≥1) (since each hour is independent of the other)

= 0.5488^2*(1-0.8781)\\\\=0.2645

(b) Find the probability that the total time he spends fishing is between two and five hours.

Prob that he does not get fish in I two hours * prob he gets fish between 3 and 5 hours

=P(0)^2 *F(1)^3\\=0.5488^2*0.2645^3\\=0.00557

(c) Find the expected number offish that he catches.

Expected value in Geometric distribution = \frac{1-p}{p}, where p = prob of getting 1 fish in one hour

= \frac{0.6}{1-0.6} \\=3

(d) Find the expected total fishing time, given that he has been fishing for four hours.

= Expected fishing time total/expected fishing time for 4 hours

=3/0.6*4

= 1.25 hours

4 0
2 years ago
A sample of baseball games shows that the mean length of the games is 179.83 minutes. (The data is given in BaseballTimes if you
Dmitriy789 [7]

Answer: Yes, This sample provide evidence that the length of time for baseball games is more than 170 minutes.

Step-by-step explanation:

Since we have given that

Mean length of the games = 179.83

Standard error = 3.75

We need to find the value of P(X>170).

So, it becomes,

z=\dfrac{X-\mu}{s}=\dfrac{179-170}{3.75}\\\\z=2.4

Let z_{crict}=1.64

Since z is greater than z(critical value).

So,Yes, This sample provide evidence that the length of time for baseball games is more than 170 minutes.

5 0
2 years ago
A supermarket has two customers waiting to pay for their purchases at counter I and one customer waiting to pay at counter II. L
Pachacha [2.7K]

Answer:

b. 0.864

Step-by-step explanation:

Let's start defining the random variables.

Y1 : ''Number of customers who spend more than $50 on groceries at counter 1''

Y2 : ''Number of customers who spend more than $50 on groceries at counter 2''

If X is a binomial random variable, the probability function for X is :

P(X=x)=(nCx)p^{x}(1-p)^{n-x}

Where P(X=x) is the probability of the random variable X to assume the value x

nCx is the combinatorial number define as :

nCx=\frac{n!}{x!(n-x)!}

n is the number of independent Bernoulli experiments taking place

And p is the success probability.

In counter I :

Y1 ~ Bi (n,p)

Y1 ~ Bi(2,0.2)

P(Y1=y1)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}

With y1 ∈ {0,1,2}

And P( Y1 = y1 ) = 0 with y1 ∉ {0,1,2}

In counter II :

Y2 ~ Bi (n,p)

Y2 ~ Bi (1,0.3)

P(Y2=y2)=(1Cy2)(0.3)^{y2}(0.7)^{1-y2}

With y2 ∈ {0,1}

And P( Y2 = y2 ) = 0 with y2 ∉ {0,1}

(1Cy2) with y2 = 0 and y2 = 1 is equal to 1 so the probability function for Y2 is :

P(Y2=y2)=(0.3)^{y2}(0.7)^{1-y2}

Y1 and Y2 are independent so the joint probability distribution is the product of the Y1 probability function and the Y2 probability function.

P(Y1=y1,Y2=y2)=P(Y1=y1).P(Y2=y2)

P(Y1=y1,Y2=y2)=(2Cy1)(0.2)^{y1}(0.8)^{2-y1}(0.3)^{y2}(0.7)^{1-y2}

With y1 ∈ {0,1,2} and y2 ∈ {0,1}

P( Y1 = y1 , Y2 = y2) = 0 when y1 ∉ {0,1,2} or y2 ∉ {0,1}

b. Not more than one of three customers will spend more than $50 can mathematically be expressed as :

Y1 + Y2 \leq 1

Y1 + Y2\leq 1 when Y1 = 0 and Y2 = 0 , when Y1 = 1 and Y2 = 0 and finally when Y1 = 0 and Y2 = 1

To calculate P(Y1+Y2\leq 1) we must sume all the probabilities that satisfy the equation :

P(Y1+Y2\leq 1)=P(Y1=0,Y2=0)+P(Y1=1,Y2=0)+P(Y1=0,Y2=1)

P(Y1=0,Y2=0)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{0}(0.7)^{1-0}=(0.8)^{2}(0.7)=0.448

P(Y1=1,Y2=0)=(2C1)(0.2)^{1}(0.8)^{2-1}(0.3)^{0}(0.7)^{1-0}=2(0.2)(0.8)(0.7)=0.224

P(Y1=0,Y2=1)=(2C0)(0.2)^{0}(0.8)^{2-0}(0.3)^{1}(0.7)^{1-1}=(0.8)^{2}(0.3)=0.192

P(Y1+Y2\leq 1)=0.448+0.224+0.192=0.864\\P(Y1+Y2\leq 1)=0.864

7 0
2 years ago
What is 5×20+2raised to the power 2​
Neporo4naja [7]

Answer:

10,404

Step-by-step explanation:

5x20+2=102

102^2=102x102=10,404

7 0
2 years ago
Point E belongs to diagonal AC of parallelogram ABCD (labeled counterclockwise) so that AE:EC=2:1. Line BE intersects CD at poin
Svetach [21]

Consider two triangles AEB and CEF. In these triangles:

  1. ∠AEB≅∠CEF (as vertical angles);
  2. ∠EAB≅∠ECF (lines AB and CD are parallel and AC is transversal, then these angles are congruent as alternate interior angles);
  3. ∠ABE≅∠CFE (lines AB and CD are parallel, BF is transversal, then these angles congruent as alternate interior angles).

Thus, by AAA theorem, \triangle AEB\sim \triangle CEF.

Corresponding sides of similar triangles are proportional, then

\dfrac{AB}{CF}=\dfrac{AE}{EC},\\ \\\dfrac{AB}{CF}=\dfrac{2}{1},\\ \\AB=2CF.

ABCD is a parallelogram, then AB=CD.

Therefore,

CD=2CF and CD=DF+CF. Equate these two expressions:

DF+CF=2CF,

DF=CF.

This gives you that DF:CF=1:1.

Answer: 1:1


7 0
2 years ago
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