Please help me!!

Question

A manufacturer knows that their items have a normally distributed length, with a mean of 17.3 inches, and standard deviation of 4.7 inches.

If 10 items are chosen at random, what is the probability that their mean length is less than 15.9 inches?

Round to 4 decimal places.

Answer 1

So, what you need to perform is a Z-test:

Z = (x̄ - μ₀) / (σ / √N)

Z = (15.9 - 17.3) / (4.7 / √10)

Z = -0.941955

Next, you want to refer to a Z-table to determine the probability:

Probability @ Z = -0.94: 0.17361

Probability @ Z = -0.95: 0.17106

Given that our Z score is between these two Z values, the probability is somewhere between the two reference probabilities. By convention, you can take the average of these two:

(0.17361 + 0.17106) / 2 = 0.172335

Which means that the mean length of these 10 items will have about a 17.23% chance of being less than 15.9 inches.

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A more accurate method would be to take a weighted average of the two values:

Using the digits past -0.94: 19.55% + 80.45% = 100%

0.17361 * 0.8045 + 0.17106 * 0.1955 = 0.173111475 ∼ 0.1731

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Checking our answer against the exact value on a TI-calculator:

STAT → TESTS → Z-Test

Input: Stats

μ₀: 17.3

σ: 4.7

x̄: 15.9

n: 10

μ: <μ₀

Calculate

p = 0.173111475 ∼ 0.1731

The averaging method is the least accurate, but fairly close to the exact. Alternatively you could just report the range: between 0.1711 and 0.1736. The weighted average is extremely close to the real, so much so that we couldn't observe the difference between them. It's a tiny bit more complicated, though. Naturally, the calculator option is the most accurate. Do what you feel comfortable with and what is allowable on an exam.