Question

(1)Given that x is a normal variable with mean μ = 52 and standard deviation σ = 6.9, find the following probabilities. (Round your answers to four decimal places.) (a) P(x ≤ 60) (b) P(x ≥ 50) (c) P(50 ≤ x ≤ 60) (2) Find z such that 15% of the area under the standard normal curve lies to the right of z. (Round your answer to two decimal places.) (3) The University of Montana ski team has six entrants in a men's downhill ski event. The coach would like the first, second, and third places to go to the team members. In how many ways can the six team entrants achieve first, second, and third places?

Answer #1

Question 1)

Answer) Standard formula for z is given by

Z= (x-μ)/σ

Therefore

(a) P(x60)

Z(60) =(60-52)/6.9

Therefore P(x60) = P(Z (60-52)/6.9)

P(x60) = P(Z1.1594)

=0.8770 (From z table)

(b) P(X≥50) = P(Z≥50)

Z(50) = (50-52)/6.9 = -0.2898

P(Z≥50) = 1-P(Z≤50)

Therefore, P(X≥50) = 1-0.3859(from z table)

Therefore, P(X≥50) = 0.6141

(c) P(50 ≤ x ≤ 60)

So for x = 50

(50-52)/6.9 =-0.2898

and for X=60

(60-52)/6.9 = 1.1594

so, P(50 ≤ x ≤ 60) = P(-0.2898≤ z ≤ 1.1594)

from z table for -0.2898 = 0.6141 (follow the steps of (b))

from z table for 1.1594 = 0.8770 (follow the steps of (a))

P(50 ≤ x ≤ 60) = Probability of x less than 60 - Probability of x greater than 50

0.8770-0.6141

= 0.2629

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