You may need to use the e table in the Acpendix of Tables to answer this questio

Question

You may need to use the e table in the Acpendix of Tables to answer this question. Need Help? Spray drift is a constant c for pesticide relationship between droplet sine and drift potential is well known. The paper "Effects of 2,4 D F 150 ym was a reasonable model for droplet sine for mater (the "control treatment") sorayed through a 760 ml/min nozzle on spray atomization, A figure in a paper supgested the normal distribution with mean 1050 pm and standard ess than 1500 m your answers to four decimal places) 950 and 1500 pm? (Round your ansmer to four decimal places) (c)How would you diara terte the smale t 2% of all The smalest 2% of dragits arc Bose s than (4) It the table in the Appendix of Tables to Home End 2 1 8 6

Explanation / Answer

A) P(x < 1500)

= P((x - mean)/SD < (1500 - mean)/SD)

= P(Z < (1500 - 1050)/150)

= P(Z < 3)

= 0.9987

P(x > 950)

= P((x - mean)/SD > (950 - mean)/SD)

= P(Z > (950 - 1050)/150)

= 1 - P(Z < -0.67)

= 1 - 0.2514

= 0.7486

B) P(950 < x < 1500)

= P((950 - mean)/SD) < (x - mean)/SD < (1500 - mean)/SD)

= P((950 - 1050)/150 < Z < (1500 - 1050)/150)

= P(-0.67 < Z < 3)

= P(Z < 3) - P(Z < -0.67)

= 0.9987 - 0.2514

= 0.7473

C) P(X < x) = 0.02

Or, P((X - mean)/SD < (x - mean)/SD) = 0.02

Or, P(Z < (x - 1050)/150) = 0.02

Or, (x - 1050)/150 = -2.05

Or, x = -2.05 * 150 + 1050

Or, x = 742.5

D) P(X > 1500)

= P((x - mean)/SD > (1500 - mean)/SD)

= P(Z > (1500 - 1050)/150)

= P(Z > 3)

= 1 - P(Z < 3)

= 1 - 0.9987

= 0.0013

This is a binomial distribution

n = 5

P = 0.0013

P(X > 1)

= 1 - P(X < 1)

= 1 - P(x = 0)

= 1 - 5C0 * (0.0013)^0 * (1 - 0.0013)^5)

= 1 - 0.9935

= 0.0065

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