We can solve each part step by step.
(a) Probability that one item's length is less than 16.4 inches.
The formula for the Z-score is given by:
\[ Z = \frac{X - \mu}{\sigma} \]
where:
- \(X = 16.4\) (the value we're interested in),
- \(\mu = 18.6\) (the mean),
- \(\sigma = 5.8\) (the standard deviation).
Substituting the values in:
\[ Z = \frac{16.4 - 18.6}{5.8} = \frac{-2.2}{5.8} \approx -0.3793 \]
Next, we need to find the probability corresponding to \(Z = -0.3793\). We can use the standard normal distribution table or a calculator for this:
\[ P(Z < -0.3793) \approx 0.3548 \] (Rounded to four decimal places)
Answer for (a): \(0.3548\)
(b) Mean and standard deviation of the sample mean length of 24 items.
For the sample mean, the mean (\(\mu_{\overline{X}}\)) is the same as the mean of the population:
\[ \mu_{\overline{X}} = \mu = 18.6 \text{ inches} \]
The standard deviation of the sample mean (\(\sigma_{\overline{X}}\)) is calculated as follows:
\[ \sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} \]
where \(n = 24\):
\[ \sigma_{\overline{X}} = \frac{5.8}{\sqrt{24}} \approx \frac{5.8}{4.899} \approx 1.1826 \]
Mean = \(18.6\) inches
Standard Deviation ≈ \(1.1826\) inches
(c) Probability that the mean length of 24 items is less than 16.4 inches.
To find this, we first calculate the Z-score for the sample mean:
\[ Z = \frac{\overline{X} - \mu_{\overline{X}}}{\sigma_{\overline{X}}} \]
where:
- \(\overline{X} = 16.4\),
- \(\mu_{\overline{X}} = 18.6\),
- \(\sigma_{\overline{X}} \approx 1.1826\).
Substituting the values in:
\[ Z = \frac{16.4 - 18.6}{1.1826} = \frac{-2.2}{1.1826} \approx -1.8645 \]
Now, we look up the probability for \(Z = -1.8645\):
\[ P(Z < -1.8645) \approx 0.0302 \] (Rounded to four decimal places)
Answer for (c): \(0.0302\)
Summary of Answers:
- (a) \(0.3548\)
- (b) Mean = \(18.6\), Standard Deviation ≈ \(1.1826\)
- (c) \(0.0302\)