Question

2) A happiness survey with normally distributed scores with mean 5.5 and standard deviation 2.3 was administered to IT workers in healthcare. a) Find the probability that a randomly selected participant’s response was greater than 5.



b) Find the probability that a randomly selected participant’s response was between 4.5 and 6.5.



c) Find the probability that the mean of a sample of 16 selected participant’s response was between 4.5 and 6.5.

Answer 1

(a)

= 5.5

= 2.3

To find P(X>5):

Z = (5 - 5.5)/2.3

= - 0.2174

By Technology, Cumulative Area Under Standard Normal Curve = 0.4139

So,

the probability that a randomly selected participant’s response was greater than 5. = 1 - 0.4139 = 0.5861

So,

Answer is:

0.5861

(b)

To find P(4.5 <X < 6.5):

For X = 4.5:

Z = (4.5 - 5.5)/2.3

= - 0.4348

By Technology, Cumulative Area Under Standard Normal Curve = 0.3319

For X = 6.5:

Z = (6.5 - 5.5)/2.3

= 0.4348

By Technology, Cumulative Area Under Standard Normal Curve = 0.6681

So,

the probability that a randomly selected participant’s response was between 4.5 and 6.5 =0.6681 - 0.3319 = 0.3362

So,

Answer is:

0.3362

(c)

n = 16

SE = 2.3/

= 0.575

To find P(4.5 < < 6.5):

For = 4.5:

Z = (4.5 - 5.5)/0.575

= - 1.7391

By Technology, Cumulative Area Under Standard Normal Curve = 0.0410

For X = 6.5:

Z = (6.5 - 5.5)/0.575

= 1.7391

By Technology, Cumulative Area Under Standard Normal Curve = 0.9590

So,

the probability that the mean of a sample of 16 selected participant’s response was between 4.5 and 6.5. = 0.9590 - 0.0410 = 0.9180

So,

Answer is:

0.9180