MATHEMATICS HIGH SCHOOL

ACT mathematics score for a particular year are normally distributed with a mean of 28 and a standard deviation of 2.4 pointsA. What is the probability a randomly selected score is greater than 30.4?

B. what is the probability a randomly selected score is less than 32.8?

C. What is the probability a randomly selected score is Between 25.6 and 32.8?

Answers

Answer 1
Answer:

Answer:

a.   0.1587

b.  0.8849

c.  0.1814

Step-by-step explanation:

a. Given that \mu=28, \ \ \sigma=2.4

-The probability a randomly selected score is greater than 30.4 is calculated as:

P(X>30.4)=1-P(X<30.4)\n\nz=(\bar X-\mu)/(\sigma)\n\n=(30.4-28)/(2.4)=1\n\n\therefore P(X>30.4)=1-P(z<1)\n\n=1-0.84134\n\n=0.1587

Hence, the probability of a score greater than 30.4 is 0.1587

b. Given that \mu=28, \ \ \ \sigma=2.4

The probability a randomly selected score is less than 32.8 is calculated as:

P(X<32.8)=P(z<(\bar X-\mu)/(\sigma))\n\nz=(\bar X-\mu)/(\sigma)\n\n=(32.8-28)/(2.4)=1.2\n\nP(X<32.8)=P(z<1.2)\n\n=0.88493

Hence, the probability that a randomly selected score is less than 32.8 is 0.8849

c. The probability that a score is between 25.6 and 32.8 is calculated as follows:

P(25.6<X<32.8)=P((\bar X-\mu)/(\sigma)<z<(\bar X-\mu)/(\sigma))\n\n=P((25.6-28)/(2.4)<z<(32.8-28)/(2.4))\n\n=P(-1<z<2.0)\n\n=0.15866+(1-0.97725)\n\n=0.1814

Hence, the probability is 0.1814

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от
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3
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Answers

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Answers

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c = √(90)

Step-by-step explanation:

Using Pythagoras' identity in the right triangle.

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A stadium has 55,000 seats. Seats sell for ​$28 in Section​ A, ​$16 in Section​ B, and ​$12 in Section C. The number of seats in Section A equals the total number of seats in Sections B and C. Suppose the stadium takes in ​$1,158,000 from each​ sold-out event. How many seats does each section​ hold?

Answers

A=number of seats in section A
B=number of seats in section B
C=number of seats in section C

We can suggest this system of equations:

A+B+C=55,000
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28A+16B+12C=1,158,000

We solve this system of equations by Gauss Method.

1             1              1               55,000
1            -1            -1                         0
28         16           12          1,158,000 


1             1              1               55,000
0            -2             -2             -55,000                                    (R₂-R₁)
0           12             16           382,000                                    (28R₁-R₂) 


1            1                1              55,000
0           -2              -2             -55,000
0            0                4               52,000                                     (6R₂+R₃)

Therefore:

4C=52,000
C=52,000/4
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-2B-2(13,000)=-55,000
-2B-26,000=-55,000
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-2B=-29,000
B=-29,000 / -2
B=14,500.

A + 14,500+13,000=55,000
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A=55,000-27,500
A=27,500.

Answer: there are 27,500 seats in section A, 14,500 seats in section B and 13,000 seats in section C.
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