..ooo Verizon 12:52 AM Page 2 the apprepnite wea e the curve Show how you armved

Question

..ooo Verizon 12:52 AM Page 2 the apprepnite wea e the curve Show how you armved at each answer. 1. A male freshman 74 inches tll is at the th percentile The amount of heating oll used ayby Ohio housholds follows a normal distribution with mean What penentage would you expect to use between 180 and 272 galyear? 200 h 3. An inteligence Quotient 0) test has noemally distributed results, wilth mean 100 and standard deviation 16. What pecentage of test takers will score between 70 and 121 b. The person that scored 810n the test is at the .th percentle 4. Chaillenge: I want to score better than 90% of all test takers. What Q score will I need?

Explanation / Answer

Quesrtion one is not visible correct. I will solve rest of the questions

QUestion 2.

Mean Heating oil used = 200 gallons

Standard deviation of oil used = 40 gallons

Let X is heating oil used by a random household

Pr(X > 150; 200; 40) = ?

Z = (150 - 200)/ 40 = -1.25

from Z - table

Pr(X > 150; 200; 40) = Pr(Z > -1.25) = 1 - Pr(Z < -1.25)

= 1 - 0.1056 = 0.8944

2(b) Pr (180 gal < X < 272 gal ; 200 gal ; 40 gal) = Pr(X < 272 gal ; 200 gal ; 40 gal) - Pr(X < 180 gal ; 200 gal ; 40 gal)

= (Z2 ) - (Z1)

where Z2 = (272 - 200)/40 = 1.8

Z1 = *180 - 200)/40 = -0.5

where is cumulative standard normal distribution.

Calculting this value from Z table

Pr (180 gal < X < 272 gal ; 200 gal ; 40 gal) = (1.8) - (-0.5) = 0.9641 - 0.3085 = 0.6556

Q.3 Mean = 100

standard deviation = 16

Let X is the IQ of a random test taker

Pr(70 < X < 121) = NORM(70 < X < 121 ; 100 ; 16) = Pr(X < 121; 100; 16) - Pr(X < 70 ; 100; 16)

= (Z2 ) - (Z1)

where Z2 = (121 - 100)/16 = 1.3125

Z1 = (70 - 100)/16 = -1.875

where is cumulative standard normal distribution.

Calculting this value from Z table

Pr(70 < X < 121) = (Z2 ) - (Z1) = (1.3125) - (-1.875) = 0.9054 - 0.0304 = 0.875

3b. If X = 81 then

Pr(X < 81; 100; 16) = (Z)

Z = (81 - 100)/16 = -1.1875

Pr(X < 81; 100; 16) = (Z) = (-1.875) = 0.12

so the percentile is 12% percentile

Q.4

Here i have to score 90% of all test takers.

so here p = 0.90

Pr(X < x ; 100; 16) = 0.90  

Z - value for p = 0.90

Z = 1.645

(X - 100)/16 = 1.645

X = 100 + 16 * 1.645 = 126.31

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.