Question

A normal population has mean μ = 31 and standard deviation σ = 6. (a) What proportion of the population is between 19 and 28? (b) What is the probability that a randomly chosen value will be between 26 and 36? Round the answers to at least four decimal places. Part 1 of 2 The proportion of the population between 19 and 28 is . Part 2 of 2 the probability that a randomly chosen value will be between 26 and 36 is?

Answer 1

Solution :

(a)

P(19 < x < 28) = P[(19 - 31)/ 6) < (x - ) /  < (28 - 31 ) / 6) ]

= P(1.33 < z < -0.5)

= P(z < -0.5) - P(z < 1.33)

= 0.5997

Proportion = 0.5997

(b)

(26 < x < 36) = P[(26 - 31)/ 6) < (x - ) /  < (36 - 31 ) / 6) ]

= P(-0.83 < z < 0.83)

= P(z < 0.83) - P(z < -0.83)

= 0.5935

Proportion = 0.5935