Logo

AskSia

Plus

Solve question 5,6,7,8 Skill- theet [7][7] Exercise 15C Note that many of the rea...
May 12, 2024
Solve questions 5,6,7,8 with answers and formulas
Solution by Steps
step 1
To solve question 5, we need to find the probability that the difference in scores between the two students is more than 7500. Since both scores are normally distributed with the same mean μ=25000\mu = 25000 and standard deviation σ=3000\sigma = 3000, the difference in scores will also be normally distributed with mean μdiff=0\mu_{diff} = 0 and standard deviation σdiff=2⋅σ=2⋅3000\sigma_{diff} = \sqrt{2} \cdot \sigma = \sqrt{2} \cdot 3000
step 2
We convert the difference of 7500 to a z-score by dividing by the standard deviation of the difference: z=7500σdiff=75002⋅3000z = \frac{7500}{\sigma_{diff}} = \frac{7500}{\sqrt{2} \cdot 3000}
step 3
Calculate the z-score: z=75002⋅3000≈1.77z = \frac{7500}{\sqrt{2} \cdot 3000} \approx 1.77
step 4
Use the standard normal distribution table or a calculator to find the probability that a z-score is greater than 1.77. This gives us the probability that one score is more than 7500 points higher than the other
step 5
The probability from the z-table or calculator for z > 1.77 is approximately 0.0384. Since we are looking for the absolute difference being more than 7500, we need to consider both tails of the distribution, effectively doubling this probability
[question 5 Answer]
The probability that the students' scores will differ by more than 7500 points is approximately 0.0768.
Key Concept
Difference of two independent normal distributions
Explanation
The difference of two independent normally distributed variables is also normally distributed, with mean equal to the difference of the means and variance equal to the sum of the variances.
step 1
For question 6, we first find the distribution of the total weight of a bag of six bananas. The mean total weight μtotal\mu_{total} is 6×180g=1080g6 \times 180g = 1080g, and the standard deviation σtotal\sigma_{total} is 6×20g\sqrt{6} \times 20g since the weights are independent
step 2
Convert the weight limit of 1 kg (1000g) to a z-score: z=1000−μtotalσtotal=1000−10806×20z = \frac{1000 - \mu_{total}}{\sigma_{total}} = \frac{1000 - 1080}{\sqrt{6} \times 20}
step 3
Calculate the z-score: z=1000−10806×20≈−1.633z = \frac{1000 - 1080}{\sqrt{6} \times 20} \approx -1.633
step 4
Use the standard normal distribution table or a calculator to find the probability that a z-score is less than -1.633
[question 6 Answer]
The probability that the weight of a randomly chosen bag of six bananas is less than 1 kg is approximately 0.0514.
Key Concept
Sum of independent normal distributions
Explanation
The sum of independent normally distributed variables is also normally distributed, with mean equal to the sum of the means and variance equal to the sum of the variances.
step 1
For question 7, we need to find the maximum number of people nn that can be at least 99% sure not to exceed the elevator capacity of 680 kg. The total weight of nn people is normally distributed with mean n×82n \times 82 kg and standard deviation n×9\sqrt{n} \times 9 kg
step 2
We use the z-score corresponding to the 99th percentile, which is approximately 2.33. We set up the inequality n×82+2.33×n×9≤680n \times 82 + 2.33 \times \sqrt{n} \times 9 \leq 680
step 3
Solve the inequality for nn. This may require numerical methods or trial and error, as it is a nonlinear equation
[question 7 Answer]
The maximum number of people who can get into the elevator to be at least 99% sure not to exceed the capacity is 7.
Key Concept
Normal distribution and confidence intervals
Explanation
To ensure a certain confidence level for a normally distributed variable, we use the z-score that corresponds to the desired percentile and set up an inequality to solve for the unknown.
step 1
For question 8, we need to find the probability that the alarm system is still working after 145 hours. The total life of the 20 batteries is normally distributed with mean 20×720 \times 7 hours and standard deviation 20×0.5\sqrt{20} \times 0.5 hours
step 2
Convert 145 hours to a z-score: z=145−(20×7)20×0.5z = \frac{145 - (20 \times 7)}{\sqrt{20} \times 0.5}
step 3
Calculate the z-score: z=145−14020×0.5≈1.118z = \frac{145 - 140}{\sqrt{20} \times 0.5} \approx 1.118
step 4
Use the standard normal distribution table or a calculator to find the probability that a z-score is greater than 1.118
[question 8 Answer]
The probability that the alarm system is still working after 145 hours is approximately 0.1319.
Key Concept
Sum of independent normal distributions and their properties
Explanation
The total life of independent batteries is normally distributed, and we use the z-score to find the probability of the system lasting a certain amount of time.
© 2023 AskSia.AI all rights reserved