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Answer :
To determine the probability that the average weight of a sample of 20 people is between 70 kg and 84 kg, we can use the properties of the normal distribution. Since the population is normally distributed, the sampling distribution of the sample mean will also be normally distributed.
Identify Given Information:
- Population mean ([tex]\mu[/tex]) = 75 kg
- Population standard deviation ([tex]\sigma[/tex]) = 15 kg
- Sample size ([tex]n[/tex]) = 20
Calculate the Standard Error of the Mean (SEM):
The standard error of the mean is calculated as follows:
[tex]SEM = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{20}} \approx 3.35[/tex]Convert the Sample Means to Z-scores:
To find probabilities using the normal distribution, convert the sample means (70 kg and 84 kg) to Z-scores using the formula:
[tex]Z = \frac{X - \mu}{SEM}[/tex]
For 70 kg:
[tex]Z_{70} = \frac{70 - 75}{3.35} \approx -1.49[/tex]
For 84 kg:
[tex]Z_{84} = \frac{84 - 75}{3.35} \approx 2.69[/tex]Find the Probability between the Z-scores:
Use standard normal distribution tables or a calculator to find the probability that [tex]Z[/tex] falls between -1.49 and 2.69.- Probability ([tex]Z < 2.69[/tex]) [tex]\approx 0.9964[/tex]
- Probability ([tex]Z < -1.49[/tex]) [tex]\approx 0.0681[/tex]
The probability that the sample mean is between 70 kg and 84 kg is:
[tex]P(70 < \bar{X} < 84) = P(Z < 2.69) - P(Z < -1.49) = 0.9964 - 0.0681 = 0.9283[/tex]Interpretation:
There is approximately a 92.83% probability that the average weight of the sample is between 70 kg and 84 kg. This high probability indicates that it is very likely for the sample mean to fall within this range, given the normal distribution of the population.
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