- Î§ = amount of change students carry
- Î§ ~ E(0.88, 0.88)
- $\stackrel{\xc2\xaf}{X}$ = average amount of change carried by a sample of 25 sstudents.
- $\stackrel{\xc2\xaf}{X}$ ~ N(0.88, 0.176)
- 0.0819
- 0.1882
- The distributions are different. Part a is exponential and part b is normal.
- length of time for an individual to complete IRS form 1040, in hours.
- mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours.
- N$\left(\text{10}\text{.53,\xc2}\frac{1}{3}\right)$
- Yes. I would be surprised, because the probability is almost 0.
- No. I would not be totally surprised because the probability is 0.2312
- the length of a song, in minutes, in the collection
- U(2, 3.5)
- the average length, in minutes, of the songs from a sample of five albums from the collection
- N(2.75, 0.0660)
- 2.71 minutes
- 0.09 minutes
- True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.
- True. According to the Central Limit Theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.
- The standard deviation of the sampling distribution of the means will decrease making it approximately the same as the standard deviation of X as the sample size increases.
- X = the yearly income of someone in a third world country
- the average salary from samples of 1,000 residents of a third world country
- $\stackrel{\xc2\xaf}{X}$ âˆ¼ N$\left(\text{2000,\xc2}\frac{\text{8000}}{\sqrt{\text{1000}}}\right)$
- Very wide differences in data values can have averages smaller than standard deviations.
- The distribution of the sample mean will have higher probabilities closer to the population mean.
P(2000 < $\stackrel{\xc2\xaf}{X}$ < 2100) = 0.1537
P(2100 < $\stackrel{\xc2\xaf}{X}$ < 2200) = 0.1317
- the total length of time for nine criminal trials
- N(189, 21)
- 0.0432
- 162.09; ninety percent of the total nine trials of this type will last 162 days or more.
- X = the salary of one elementary school teacher in the district
- X ~ N(44,000, 6,500)
- Î£X ~ sum of the salaries of ten elementary school teachers in the sample
- Î£X ~ N(44000, 20554.80)
- 0.9742
- $52,330.09
- 466,342.04
- Sampling 70 teachers instead of ten would cause the distribution to be more spread out. It would be a more symmetrical normal curve.
- If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.
- X = the closing stock prices for U.S. semiconductor manufacturers
- i. $20.71; ii. $17.31; iii. 35
- Exponential distribution, Î§ ~ Exp$\left(\frac{1}{20.71}\right)$
- Answers will vary.
- i. $20.71; ii. $11.14
- Answers will vary.
- Answers will vary.
- Answers will vary.
- N$\left(\text{20}\text{.71,\xc2}\frac{17.31}{\sqrt{5}}\right)$
- Check studentâ€™s solution.
- $\stackrel{\xc2\xaf}{X}$ ~ N$\left(\text{60,\xc2}\frac{9}{\sqrt{25}}\right)$
- 0.5000
- 59.06
- 0.8536
- 0.1333
- N(1500, 45)
- 1530.35
- 0.6877
- We have Î¼ = 17, Ïƒ = 0.8, $\stackrel{\xc2\xaf}{x}$ = 16.7, and n = 30. To calculate the probability, we use
normalcdf
(lower, upper, Î¼, $\frac{\mathrm{\xcf\u0192}}{\sqrt{n}}$) =normalcdf
$\left(E\xe2\u20ac\u201c\text{99,16}\text{.7,17,}\frac{0.\text{8}}{\sqrt{\text{30}}}\right)$ = 0.0200. - If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 lifetime hours is only 2%. Therefore, the class was justified to question the claim.
- For the sample, we have n = 100, $\stackrel{\xc2\xaf}{x}$ = 0.862, s = 0.05
- $\mathrm{\xce\pounds}\stackrel{\xc2\xaf}{x}$ = 85.65, Î£s = 5.18
normalcdf
(396.9,E99,(465)(0.8565),(0.05)($\sqrt{465}$)) â‰ˆ 1- Since the probability of a sample of size 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that Mars is correctly labeling their M&M packages.
Use normalcdf
$\left(E\xe2\u20ac\u201c\text{99,1}\text{.1,1,}\frac{1}{\sqrt{\text{70}}}\right)$ = 0.7986. This means that there is an 80% chance that the service time will be less than 1.1 hours. It could be wise to schedule more time since there is an associated 20% chance that the maintenance time will be greater than 1.1 hours.
We assume that the weights of coins are normally distributed in the population. Since we have normalcdf
$\left(5.\text{111,5}\text{.291,5}\text{.201,}\frac{0.\text{065}}{\sqrt{\text{280}}}\right)$ â‰ˆ 0.8338, we expect (1 â€“ 0.8338)280 â‰ˆ 47 coins to be rejected.