Question

A population of values has a normal distribution with μ=74.4μ=74.4 and σ=34.9σ=34.9. You intend to draw a random sample of size n=29n=29.

Find the probability that a single randomly selected value is between 56.9 and 89.3.
P(56.9 < X < 89.3) =

Find the probability that a sample of size n=29n=29 is randomly selected with a mean between 56.9 and 89.3.
P(56.9 < M < 89.3) =

Enter your answers as numbers accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution :

Given that ,

mean = = 74.4

standard deviation = = 34.9

P(56.9 < x < 89.3) = P((56.9 - 74.4)/ 34.9) < (x - ) /  < (89.3 - 74.4) / 34.9) )

= P(-0.501 < z < 0.427)

= P(z < 0.427) - P(z < -0.501)

= 0.6653 - 0.3082

= 0.3571

Probability = 0.3571

(b)

n = 29

M = 74.4

M = / n = 34.9 / 29 = 6.4808

P(56.9 < M < 89.3) = P((56.9 - 74.4) / 6.4808 <( - M) / M < (89.3 - 74.4) / 6.4808))

= P(-2.70 < Z < 2.299)

= P(Z < 2.299) - P(Z < -2.700)

= 0.9892 - 0.0035

= 0.9857

Probability = 0.9857