Question

A population of values has a normal distribution with μ = 127.5 and σ = 96.9 . You intend to draw a random sample of size n = 150 .

Find the probability that a single randomly selected value is greater than 114.8. P(x > 114.8) =

Find the probability that a sample of size n = 150 is randomly selected with a mean greater than 114.8. P( ¯ x > 114.8) =

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 = = 127.5

standard deviation = = 96.9

P(x > 114.8) = 1 - P(x < 114.8)

= 1 - P((x - ) / < (114.8 - 127.5) / 96.9)

= 1 - P(z < -0.131)

= 1 - 0.4479 Using standard normal table.

= 0.5521

The probability that a single randomly selected value is greater than 114.8 is 0.5521

n = 150

=   = 127.5 and

= / n = 96.9 / 150 = 7.9119

P( > 114.8) = 1 - P( < 114.8)

= 1 - P(( - ) / < (114.8 - 127.5) / 7.9119)

= 1 - P(z < -1.764)

= 1 - 0.0389 Using standard normal table.

= 0.9611

The probability that a sample of size n = 150 is randomly selected with a mean greater than 114.8 is 0.9611