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Sagot :
Let's go through each part of the question, step by step.
### Part (a)
Question: What is the distribution of [tex]\( X \)[/tex] ?
The distribution of [tex]\( X \)[/tex] is given by mean [tex]\( \mu = 58 \)[/tex] mL and standard deviation [tex]\( \sigma = 11 \)[/tex] mL. This can be written as:
[tex]\[ X \sim N (\mu, \sigma) \][/tex]
[tex]\[ X \sim N (58, 11) \][/tex]
### Part (b)
Question: What is the distribution of [tex]\( \bar{x} \)[/tex] ?
For the sample mean [tex]\( \bar{x} \)[/tex], the mean remains the same ([tex]\( \mu_{\bar{x}} = 58 \)[/tex] mL), but the standard deviation changes. The standard deviation of the sample mean is given by:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
where [tex]\( n \)[/tex] is the sample size. Here, [tex]\( n = 16 \)[/tex]:
[tex]\[ \sigma_{\bar{x}} = \frac{11}{\sqrt{16}} = \frac{11}{4} = 2.75 \][/tex]
Thus, the distribution of [tex]\( \bar{x} \)[/tex] can be written as:
[tex]\[ \bar{x} \sim N (\mu_{\bar{x}}, \sigma_{\bar{x}}) \][/tex]
[tex]\[ \bar{x} \sim N (58, 2.75) \][/tex]
### Part (c)
Question: If a single randomly selected individual is observed, find the probability that this person consumes between 62 mL and 65.5 mL.
For a single individual, we can use the standard normal distribution to find this probability. First, we convert the given values to z-scores using the transformation:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
1. For the lower bound [tex]\( 62 \)[/tex] mL:
[tex]\[ Z_{\text{lower}} = \frac{62 - 58}{11} = \frac{4}{11} \approx 0.3636 \][/tex]
2. For the upper bound [tex]\( 65.5 \)[/tex] mL:
[tex]\[ Z_{\text{upper}} = \frac{65.5 - 58}{11} = \frac{7.5}{11} \approx 0.6818 \][/tex]
Using the standard normal distribution (Z-table or a computational tool), we find the corresponding probabilities for the z-scores.
- Probability for [tex]\( Z_{\text{lower}} \)[/tex]:
[tex]\[ \Phi(0.3636) \approx 0.6411 \][/tex]
- Probability for [tex]\( Z_{\text{upper}} \)[/tex]:
[tex]\[ \Phi(0.6818) \approx 0.7420 \][/tex]
The probability of a single individual consuming between 62 mL and 65.5 mL is the difference between these two probabilities:
[tex]\[ \text{Probability} = \Phi(0.6818) - \Phi(0.3636) \approx 0.7420 - 0.6411 = 0.1104 \][/tex]
### Part (d)
Question: For the group of 16 pancake eaters, find the probability that the average amount of syrup is between 62 mL and 65.5 mL.
For the group, the sample mean follows the distribution:
[tex]\[ \bar{x} \sim N (58, 2.75) \][/tex]
We again convert the given values to z-scores using the sample mean's standard deviation.
1. For the lower bound [tex]\( 62 \)[/tex] mL:
[tex]\[ Z_{\text{lower}} = \frac{62 - 58}{2.75} = \frac{4}{2.75} \approx 1.4545 \][/tex]
2. For the upper bound [tex]\( 65.5 \)[/tex] mL:
[tex]\[ Z_{\text{upper}} = \frac{65.5 - 58}{2.75} = \frac{7.5}{2.75} \approx 2.7273 \][/tex]
Using the standard normal distribution:
- Probability for [tex]\( Z_{\text{lower}} \)[/tex]:
[tex]\[ \Phi(1.4545) \approx 0.9265 \][/tex]
- Probability for [tex]\( Z_{\text{upper}} \)[/tex]:
[tex]\[ \Phi(2.7273) \approx 0.9962 \][/tex]
The probability of the group's average consumption being between 62 mL and 65.5 mL is:
[tex]\[ \text{Probability} = \Phi(2.7273) - \Phi(1.4545) \approx 0.9962 - 0.9265 = 0.0697 \][/tex]
### Part (e)
Question: For part (d), is the assumption that the distribution is normal necessary?
Yes, the assumption that the distribution is normal is necessary. This is because the Central Limit Theorem states that for a sufficiently large sample size (generally [tex]\( n \geq 30 \)[/tex]), the sampling distribution of the sample mean will be approximately normally distributed regardless of the distribution of the population. However, for smaller sample sizes, the normality of the population distribution helps ensure the normality of the sample mean distribution. In this case, since the sample size is 16, which is not very large, assuming a normal distribution for the population ensures the validity of our calculations.
### Part (a)
Question: What is the distribution of [tex]\( X \)[/tex] ?
The distribution of [tex]\( X \)[/tex] is given by mean [tex]\( \mu = 58 \)[/tex] mL and standard deviation [tex]\( \sigma = 11 \)[/tex] mL. This can be written as:
[tex]\[ X \sim N (\mu, \sigma) \][/tex]
[tex]\[ X \sim N (58, 11) \][/tex]
### Part (b)
Question: What is the distribution of [tex]\( \bar{x} \)[/tex] ?
For the sample mean [tex]\( \bar{x} \)[/tex], the mean remains the same ([tex]\( \mu_{\bar{x}} = 58 \)[/tex] mL), but the standard deviation changes. The standard deviation of the sample mean is given by:
[tex]\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \][/tex]
where [tex]\( n \)[/tex] is the sample size. Here, [tex]\( n = 16 \)[/tex]:
[tex]\[ \sigma_{\bar{x}} = \frac{11}{\sqrt{16}} = \frac{11}{4} = 2.75 \][/tex]
Thus, the distribution of [tex]\( \bar{x} \)[/tex] can be written as:
[tex]\[ \bar{x} \sim N (\mu_{\bar{x}}, \sigma_{\bar{x}}) \][/tex]
[tex]\[ \bar{x} \sim N (58, 2.75) \][/tex]
### Part (c)
Question: If a single randomly selected individual is observed, find the probability that this person consumes between 62 mL and 65.5 mL.
For a single individual, we can use the standard normal distribution to find this probability. First, we convert the given values to z-scores using the transformation:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
1. For the lower bound [tex]\( 62 \)[/tex] mL:
[tex]\[ Z_{\text{lower}} = \frac{62 - 58}{11} = \frac{4}{11} \approx 0.3636 \][/tex]
2. For the upper bound [tex]\( 65.5 \)[/tex] mL:
[tex]\[ Z_{\text{upper}} = \frac{65.5 - 58}{11} = \frac{7.5}{11} \approx 0.6818 \][/tex]
Using the standard normal distribution (Z-table or a computational tool), we find the corresponding probabilities for the z-scores.
- Probability for [tex]\( Z_{\text{lower}} \)[/tex]:
[tex]\[ \Phi(0.3636) \approx 0.6411 \][/tex]
- Probability for [tex]\( Z_{\text{upper}} \)[/tex]:
[tex]\[ \Phi(0.6818) \approx 0.7420 \][/tex]
The probability of a single individual consuming between 62 mL and 65.5 mL is the difference between these two probabilities:
[tex]\[ \text{Probability} = \Phi(0.6818) - \Phi(0.3636) \approx 0.7420 - 0.6411 = 0.1104 \][/tex]
### Part (d)
Question: For the group of 16 pancake eaters, find the probability that the average amount of syrup is between 62 mL and 65.5 mL.
For the group, the sample mean follows the distribution:
[tex]\[ \bar{x} \sim N (58, 2.75) \][/tex]
We again convert the given values to z-scores using the sample mean's standard deviation.
1. For the lower bound [tex]\( 62 \)[/tex] mL:
[tex]\[ Z_{\text{lower}} = \frac{62 - 58}{2.75} = \frac{4}{2.75} \approx 1.4545 \][/tex]
2. For the upper bound [tex]\( 65.5 \)[/tex] mL:
[tex]\[ Z_{\text{upper}} = \frac{65.5 - 58}{2.75} = \frac{7.5}{2.75} \approx 2.7273 \][/tex]
Using the standard normal distribution:
- Probability for [tex]\( Z_{\text{lower}} \)[/tex]:
[tex]\[ \Phi(1.4545) \approx 0.9265 \][/tex]
- Probability for [tex]\( Z_{\text{upper}} \)[/tex]:
[tex]\[ \Phi(2.7273) \approx 0.9962 \][/tex]
The probability of the group's average consumption being between 62 mL and 65.5 mL is:
[tex]\[ \text{Probability} = \Phi(2.7273) - \Phi(1.4545) \approx 0.9962 - 0.9265 = 0.0697 \][/tex]
### Part (e)
Question: For part (d), is the assumption that the distribution is normal necessary?
Yes, the assumption that the distribution is normal is necessary. This is because the Central Limit Theorem states that for a sufficiently large sample size (generally [tex]\( n \geq 30 \)[/tex]), the sampling distribution of the sample mean will be approximately normally distributed regardless of the distribution of the population. However, for smaller sample sizes, the normality of the population distribution helps ensure the normality of the sample mean distribution. In this case, since the sample size is 16, which is not very large, assuming a normal distribution for the population ensures the validity of our calculations.
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