Answer :
Let's go through each part of the question step by step:
a. Distribution of [tex]\(X\)[/tex]:
The time students spend studying in the library follows a normal distribution with a mean ([tex]\( \mu \)[/tex]) of 45 minutes and a standard deviation ([tex]\( \sigma \)[/tex]) of 17 minutes. Thus, we denote this as:
[tex]\[ X \sim N(45, 17) \][/tex]
b. Distribution of [tex]\(\bar{x}\)[/tex] (sample mean):
The distribution of the sample mean [tex]\(\bar{x}\)[/tex] for 46 students will also be normal. Its mean will be the same as the mean of the individual times, 45 minutes, but the standard deviation will be the standard deviation divided by the square root of the sample size ([tex]\( n = 46 \)[/tex]):
[tex]\[ \bar{x} \sim N(45, \frac{17}{\sqrt{46}}) \][/tex]
The standard deviation of the sample mean is approximately 2.5065.
c. Distribution of [tex]\(\sum x\)[/tex] (total study time):
The total study time for 46 students will also be normally distributed. The mean will be the sum of the individual means, and the standard deviation will be the original standard deviation times the square root of the sample size:
[tex]\[ \sum x \sim N(2070, 17 \times \sqrt{46}) \][/tex]
The standard deviation of the total study time is approximately 115.2996.
d. Probability that one randomly selected student's time is between 46 and 50 minutes:
To find this probability, we calculate:
1. The z-score for 46 minutes:
[tex]\[ z = \frac{46 - 45}{17} \][/tex]
2. The z-score for 50 minutes:
[tex]\[ z = \frac{50 - 45}{17} \][/tex]
Find the probability that a z-score falls between these two values. The probability is approximately 0.0922.
e. Probability that the average time for 46 students is between 46 and 50 minutes:
1. Calculate the z-score for 46 minutes using the sample mean standard deviation:
[tex]\[ z = \frac{46 - 45}{2.5065} \][/tex]
2. Calculate the z-score for 50 minutes:
[tex]\[ z = \frac{50 - 45}{2.5065} \][/tex]
Find the probability that the z-score for the sample mean falls between these values. The probability is approximately 0.3219.
f. Probability that the total study time for the 46 students is more than 2024 minutes:
1. Calculate the z-score for 2024 minutes:
[tex]\[ z = \frac{2024 - 2070}{115.2996} \][/tex]
Find the probability that a z-score is greater than this value. The probability is approximately 0.6550.
g. Is the assumption of normality necessary for parts e and f?
No, the assumption of normality is not necessary due to the Central Limit Theorem, as the sample size is sufficiently large.
h. Least total time for the top 10% of total study time:
To find the threshold for the top 10%, we find the z-score that corresponds to the 90th percentile and calculate the total time using:
[tex]\[ \text{Total time} = z \times \text{standard deviation of total} + \text{mean of total} \][/tex]
The least total time needed to be in the top 10% is approximately 2217.7624 minutes.
a. Distribution of [tex]\(X\)[/tex]:
The time students spend studying in the library follows a normal distribution with a mean ([tex]\( \mu \)[/tex]) of 45 minutes and a standard deviation ([tex]\( \sigma \)[/tex]) of 17 minutes. Thus, we denote this as:
[tex]\[ X \sim N(45, 17) \][/tex]
b. Distribution of [tex]\(\bar{x}\)[/tex] (sample mean):
The distribution of the sample mean [tex]\(\bar{x}\)[/tex] for 46 students will also be normal. Its mean will be the same as the mean of the individual times, 45 minutes, but the standard deviation will be the standard deviation divided by the square root of the sample size ([tex]\( n = 46 \)[/tex]):
[tex]\[ \bar{x} \sim N(45, \frac{17}{\sqrt{46}}) \][/tex]
The standard deviation of the sample mean is approximately 2.5065.
c. Distribution of [tex]\(\sum x\)[/tex] (total study time):
The total study time for 46 students will also be normally distributed. The mean will be the sum of the individual means, and the standard deviation will be the original standard deviation times the square root of the sample size:
[tex]\[ \sum x \sim N(2070, 17 \times \sqrt{46}) \][/tex]
The standard deviation of the total study time is approximately 115.2996.
d. Probability that one randomly selected student's time is between 46 and 50 minutes:
To find this probability, we calculate:
1. The z-score for 46 minutes:
[tex]\[ z = \frac{46 - 45}{17} \][/tex]
2. The z-score for 50 minutes:
[tex]\[ z = \frac{50 - 45}{17} \][/tex]
Find the probability that a z-score falls between these two values. The probability is approximately 0.0922.
e. Probability that the average time for 46 students is between 46 and 50 minutes:
1. Calculate the z-score for 46 minutes using the sample mean standard deviation:
[tex]\[ z = \frac{46 - 45}{2.5065} \][/tex]
2. Calculate the z-score for 50 minutes:
[tex]\[ z = \frac{50 - 45}{2.5065} \][/tex]
Find the probability that the z-score for the sample mean falls between these values. The probability is approximately 0.3219.
f. Probability that the total study time for the 46 students is more than 2024 minutes:
1. Calculate the z-score for 2024 minutes:
[tex]\[ z = \frac{2024 - 2070}{115.2996} \][/tex]
Find the probability that a z-score is greater than this value. The probability is approximately 0.6550.
g. Is the assumption of normality necessary for parts e and f?
No, the assumption of normality is not necessary due to the Central Limit Theorem, as the sample size is sufficiently large.
h. Least total time for the top 10% of total study time:
To find the threshold for the top 10%, we find the z-score that corresponds to the 90th percentile and calculate the total time using:
[tex]\[ \text{Total time} = z \times \text{standard deviation of total} + \text{mean of total} \][/tex]
The least total time needed to be in the top 10% is approximately 2217.7624 minutes.