Question

A distribution of values is normal with a mean of 80 and a standard deviation of 18. From this distribution, you are drawing samples of size 13. Find the interval containing the middle-most 32% of sample means: Incorrect Enter your answer using interval notation. In this context, either inclusive or exclusive intervals would be acceptable. Your numbers should be accurate to 1 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Solution:-

Given that,

mean = = 80

standard deviation = = 18

n = 13

= = 80

= / n = 18 / 13 = 4.99

Using standard normal table,

P( -z < Z < z) = 32%

= P(Z < z) - P(Z <-z ) = 0.32

= 2P(Z < z) - 1 = 0.32

= 2P(Z < z) = 1 + 0.32

= P(Z < z) = 1.32 / 2

= P(Z < z) = 0.66

= P(Z < 0.412) = 0.66

= z  ± 0.412

Using z-score formula  

= z * +

= -0.412 * 4.99 + 80

= 77.9

Using z-score formula  

= z * +

= 0.412 * 4.99 + 80

= 82.1

The interval is = 77.9 to 82.1