Answer :
Certainly! Let's break down the problem step by step using the given statistical parameters:
Given:
- Population mean germination time ([tex]\(\mu\)[/tex]) = 22 days
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 10.3 days
We'll solve each part one by one.
### Part (a):
Find the probability that the population mean germination time will be less than 14 days.
1. First, we standardize the value 14 to find the corresponding z-score. The formula to find the z-score is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X = 14\)[/tex], [tex]\(\mu = 22\)[/tex], and [tex]\(\sigma = 10.3\)[/tex].
2. Plugging in the values:
[tex]\[ z = \frac{14 - 22}{10.3} = \frac{-8}{10.3} \approx -0.7767 \][/tex]
3. Using the z-score table or a standard normal distribution calculator, we find the probability corresponding to a z-score of -0.7767.
4. The cumulative distribution function (CDF) of the standard normal distribution at [tex]\(z = -0.7767\)[/tex] gives us the probability:
[tex]\[ P(X < 14) \approx 0.2187 \][/tex]
So, the probability that the population mean germination time will be less than 14 days is approximately 0.2187.
### Part (b):
Find the probability that the population mean germination time will be between 10 and 15 days.
1. We standardize both values (10 and 15) to find their corresponding z-scores.
2. For [tex]\(X = 10\)[/tex]:
[tex]\[ z_{\text{lower}} = \frac{10 - 22}{10.3} = \frac{-12}{10.3} \approx -1.1650 \][/tex]
3. For [tex]\(X = 15\)[/tex]:
[tex]\[ z_{\text{upper}} = \frac{15 - 22}{10.3} = \frac{-7}{10.3} \approx -0.6796 \][/tex]
4. Using the z-score table or a standard normal distribution calculator, we find the probabilities for these z-scores:
- [tex]\(P(Z < -1.1650) \approx 0.1222\)[/tex]
- [tex]\(P(Z < -0.6796) \approx 0.2486\)[/tex]
5. The probability that the germination time is between 10 and 15 days is the difference between these probabilities:
[tex]\[ P(10 < X < 15) = P(Z < -0.6796) - P(Z < -1.1650) \approx 0.2486 - 0.1222 = 0.1264 \][/tex]
So, the probability that the population mean germination time will be between 10 and 15 days is approximately 0.1264.
### Part (c):
Find the probability that the mean germination time of a sample of 64 seeds will be within 17 days of the population mean.
1. Standardize the deviation margin of 17 days for a sample size of 64.
2. When we're dealing with a sample mean, the standard deviation of the sample mean ([tex]\(\sigma_{\text{sample mean}}\)[/tex]) is:
[tex]\[ \sigma_{\text{sample mean}} = \frac{\sigma}{\sqrt{n}} \][/tex]
where [tex]\(\sigma = 10.3\)[/tex] and [tex]\(n = 64\)[/tex].
3. Calculate the standard deviation of the sample mean:
[tex]\[ \sigma_{\text{sample mean}} = \frac{10.3}{\sqrt{64}} = \frac{10.3}{8} = 1.2875 \][/tex]
4. The z-scores for [tex]\(\pm 17\)[/tex] days from the mean ([tex]\(\mu = 22\)[/tex]) are:
[tex]\[ z_{\text{lower}} = \frac{(22 - 17) - 22}{1.2875} = \frac{5 - 22}{1.2875} = \frac{-17}{1.2875} \approx -13.21 \][/tex]
[tex]\[ z_{\text{upper}} = \frac{(22 + 17) - 22}{1.2875} = \frac{39 - 22}{1.2875} = \frac{17}{1.2875} \approx 13.21 \][/tex]
5. Using the z-score table or a standard normal distribution calculator, we find that the CDF value for [tex]\(z = ±13.21\)[/tex] is very close to 1 for practical purposes.
6. Therefore, the probability that the sample mean will be within 17 days of the population mean is essentially 1.
So, the probability that the mean germination time of a sample of 64 seeds will be within 17 days of the population mean is approximately 1.0.
Given:
- Population mean germination time ([tex]\(\mu\)[/tex]) = 22 days
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 10.3 days
We'll solve each part one by one.
### Part (a):
Find the probability that the population mean germination time will be less than 14 days.
1. First, we standardize the value 14 to find the corresponding z-score. The formula to find the z-score is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where [tex]\(X = 14\)[/tex], [tex]\(\mu = 22\)[/tex], and [tex]\(\sigma = 10.3\)[/tex].
2. Plugging in the values:
[tex]\[ z = \frac{14 - 22}{10.3} = \frac{-8}{10.3} \approx -0.7767 \][/tex]
3. Using the z-score table or a standard normal distribution calculator, we find the probability corresponding to a z-score of -0.7767.
4. The cumulative distribution function (CDF) of the standard normal distribution at [tex]\(z = -0.7767\)[/tex] gives us the probability:
[tex]\[ P(X < 14) \approx 0.2187 \][/tex]
So, the probability that the population mean germination time will be less than 14 days is approximately 0.2187.
### Part (b):
Find the probability that the population mean germination time will be between 10 and 15 days.
1. We standardize both values (10 and 15) to find their corresponding z-scores.
2. For [tex]\(X = 10\)[/tex]:
[tex]\[ z_{\text{lower}} = \frac{10 - 22}{10.3} = \frac{-12}{10.3} \approx -1.1650 \][/tex]
3. For [tex]\(X = 15\)[/tex]:
[tex]\[ z_{\text{upper}} = \frac{15 - 22}{10.3} = \frac{-7}{10.3} \approx -0.6796 \][/tex]
4. Using the z-score table or a standard normal distribution calculator, we find the probabilities for these z-scores:
- [tex]\(P(Z < -1.1650) \approx 0.1222\)[/tex]
- [tex]\(P(Z < -0.6796) \approx 0.2486\)[/tex]
5. The probability that the germination time is between 10 and 15 days is the difference between these probabilities:
[tex]\[ P(10 < X < 15) = P(Z < -0.6796) - P(Z < -1.1650) \approx 0.2486 - 0.1222 = 0.1264 \][/tex]
So, the probability that the population mean germination time will be between 10 and 15 days is approximately 0.1264.
### Part (c):
Find the probability that the mean germination time of a sample of 64 seeds will be within 17 days of the population mean.
1. Standardize the deviation margin of 17 days for a sample size of 64.
2. When we're dealing with a sample mean, the standard deviation of the sample mean ([tex]\(\sigma_{\text{sample mean}}\)[/tex]) is:
[tex]\[ \sigma_{\text{sample mean}} = \frac{\sigma}{\sqrt{n}} \][/tex]
where [tex]\(\sigma = 10.3\)[/tex] and [tex]\(n = 64\)[/tex].
3. Calculate the standard deviation of the sample mean:
[tex]\[ \sigma_{\text{sample mean}} = \frac{10.3}{\sqrt{64}} = \frac{10.3}{8} = 1.2875 \][/tex]
4. The z-scores for [tex]\(\pm 17\)[/tex] days from the mean ([tex]\(\mu = 22\)[/tex]) are:
[tex]\[ z_{\text{lower}} = \frac{(22 - 17) - 22}{1.2875} = \frac{5 - 22}{1.2875} = \frac{-17}{1.2875} \approx -13.21 \][/tex]
[tex]\[ z_{\text{upper}} = \frac{(22 + 17) - 22}{1.2875} = \frac{39 - 22}{1.2875} = \frac{17}{1.2875} \approx 13.21 \][/tex]
5. Using the z-score table or a standard normal distribution calculator, we find that the CDF value for [tex]\(z = ±13.21\)[/tex] is very close to 1 for practical purposes.
6. Therefore, the probability that the sample mean will be within 17 days of the population mean is essentially 1.
So, the probability that the mean germination time of a sample of 64 seeds will be within 17 days of the population mean is approximately 1.0.
