please help with 16,17, and 18 Clas × Clas × KCSecure https://www.webassign.net/

Question

please help with 16,17, and 18

Clas × Clas × KCSecure https://www.webassign.net/web/Student/Assignment-Responses/submit?dep 19239788 16. + -6 points My Notes All applicants at a large university are required to take a special entrance exam before they are admitted. The exam scores are known to be normally distributed with a mean of 500 and a standard deviation of 60. Applicants must score 440 or more on the exam before they are admitted (a) What proportion of all applicants taking the exam is granted admission? (Round your answer to four decimal places.) (b) What proportion of all applicants will score 620 or higher on the exam? (Round your answer to four decimal places.) (c) For the coming academic year, 2400 applicants have registered to take the exam. How many do we expect to be qualified for admission to the university? (Round your answer to the nearest whole number.) applicants 17. -12 points My A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same diameter. The diameters have a normal distribution with a mean of 1.2 centimeters (cm) and a standard deviation of 0.02 cm. The manufacturer has determined that diameters in the range of 1.16 to 1.24 cm are acceptable. What proportion of all bearings falls in the acceptable range? (Round your answer to four decimal places.) 18. + -12 points MyN A large manufacturing plant uses lightbulbs with lifetimes that are normally distributed with a mean of 1500 hours and a standard deviation of 50 hours. To minimize the number of bulbs that burn out during operating hours, all bulbs are replaced at once. How often should the bulbs be replaced so that no more than 196 burn out between replacement periods? (Round your answer to one decimal place.) hr 12:25 PM O Type here to search 9/24/2018 2

Explanation / Answer

Question 16

Here mean examscores = 500

standard deviation = 60

(a) Here if x is the score of a random student

Pr(student is granted admission) = Pr(x > 440 ; 500 ; 60)

Z = (440 - 500)/60 = -1

Pr(student is granted admission) = Pr(x > 440 ; 500 ; 60) = 1 - Pr(Z < -1) = 1 - 0.1586 = 0.8413

(b) Pr(x > 620) = 1 - Pr(x < 620) = 1 - NORMDIST(x < 620 ; 500 ; 60)

Z = (620 - 500)/60 = 2

Pr(x > 620) = 1 - Pr(x < 620) = 1 - NORMDIST(x < 620 ; 500 ; 60) = 1 - Pr(Z < 2) = 1 - 0.9772 = 0.0228

(c) Expected students to be qualified for admission to the unifveristy = 2400 * 0.8413 = 2019 applicants

Question 18

Here mean diameter = 1.2 centimeter

standard deviation = 0.02 cm

Here

Pr(1.16 cm < x < 1.24 cm) = Pr(x < 1.24 cm ; 1.20 cm ; 0.02 cm) - Pr(x < 1.16 cm ; 1.20 cm; 0.02 cm)

Z2 = (1.24 - 1.20)/0.02 = 2

Z1 = (1.16 - 1.20)/0.02 = -2

Pr(1.16 cm < x < 1.24 cm) = Pr(x < 1.24 cm ; 1.20 cm ; 0.02 cm) - Pr(x < 1.16 cm ; 1.20 cm; 0.02 cm)

= Pr(Z <2) - Pr(Z < -2)

= 0.97725 - 0.02275 = 0.9545

Question 18

Here the mean lightbulbs = 1500 hours

standard devviation = 50 hours

Here we have to find the lower 1% so there will no more than 1% burn out between replacement periods.

Pr(x < x0 ; 1500 hours ; 50 hours) = 0.01

Z = -2.32635

(x0 - 1500)/50 = -2.32635

x0 = 1383.68 hours

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