Question

A distribution of values is normal with a mean of 200 and a standard deviation of 19. From this distribution, you are drawing samples of size 33. Find the interval containing the middle-most 86% of sample means: Enter your answer using interval notation. In this context, either inclusive [] or exclusive () intervals are acceptable. Your numbers should be accurate to 1 decimal places.

Answer 1

The  interval containing the middle-most 86% of sample mean in the interval notation is  is (196.4, 203.5)

The 50 +/- 86/2 percentile will make up the center 86% of the sample means' distribution.

The 86th percentile's z-score is 1.08

sample mean population mean = 200

standard deviation from mean = 19

sample size = 33.

Standard error (SE) is equal to 19 / sqrt(33) = 3.3077.

86th percentile x can be located using the formula z = (x - population mean)/sample error.

1.08 = (x - 200)/3.3077

x = 3.3077* 1.08 + 200 = 203.57

that is the interval's upper bound.

it is 3.5721 over the population mean (203.57 - 200).

Due to the symmetry of the normal distribution, the interval's bottom bound will be 1.66 standard deviations below the population mean (200 - 3.572 = 196.428).

then the interval is (196.4, 203.5).

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