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Review Quiz Started on Saturday, 29 March 2025, 3:27 AM State Finished Completed on Saturday, 29 March 2025, 4:05 AM Time taken 37 mins 57 secs Marks 20.00/20. Grade 100.00 out of 100. Question 1 On any given day, approximately 37.5 percent of the cars parked in the De Anza parking garage are parked crookedly. We randomly survey 22 cars. We are interested in the number of cars that are parked crookedly. What is the probability that at least 10 of the 22 cars are parked crookedly? a. 0. b. 0. c. 0. d. 0. Question 2 The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. If the mean is significantly greater than the standard deviation, which of the following statements is true? I. The data cannot follow the uniform distribution. II. The data cannot follow the exponential distribution. III. The data cannot follow the normal distribution. a. I only b. II only c. III only d. I, II, and III Question 3 A health club is interested in knowing how many times a typical member uses the club in a week. They decide to ask every tenth customer on a specified day to complete a short survey, including information about how many times they have visited the club in the past week. What kind of data is number of visits per week? a. Qualitative b. Quantitative-continuous c. Quantitative-discrete Question 4 A community college offers classes six days a week: Monday through Saturday. Maria conducted a study of the students in her classes to determine how many days per week the students who are in her classes come to campus for classes. In each of her five classes she randomly selected 10 students and asked them how many days they come to campus for classes. Each of her classes are the same size. The results of her survey are summarized in the table below:
What is the 60th^ percentile for this data? a. 2 b. 3 c. 4 d. 5 Question 5 According to a college survey, 22% of all students work full time. Find the mean for the number of students who work full time in samples of size 16. a. 3. b. 2. c. 4. d. 0. Question 6 Find the standard deviation, σ, for the binomial distribution which has the stated values of n and p. n = 38; p = 0. a. σ = 0. b. σ = 6. c. σ = 3. d. σ = 7. Question 7 The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Find the probability that it takes at least eight minutes to find a parking space. a. 0. b. 0. c. 0. d. 0. Question 8 The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Seventy percent of the time, it takes more than how many minutes to find a parking space? a. 1. b. 2. c. 3. d. 6. Question 9
a. 15 b. 2. c. 1. d. 3. Question 14 A community college offers classes six days a week: Monday through Saturday. Maria conducted a study of the students in her classes to determine how many days per week the students who are in her classes come to campus for classes. In each of her five classes she randomly selected 10 students and asked them how many days they come to campus for classes. Each of her classes are the same size. The results of her survey are summarized in the table below: X ~ U (4, 10). Find the 30th^ percentile. a. 0. b. 3 c. 5. d. 6. Question 15 The length of time it takes to find a parking space at 9 a.m. follows a normal distribution with a mean of five minutes and a standard deviation of two minutes. Based on the given information and numerically justified, would you be surprised if it took less than one minute to find a parking space? a. Yes b. No c. Unable to determine Question 16 One hundred eighteen students were asked what type of color their bedrooms were painted: light colors, dark colors, or vibrant colors. The results were tabulated according to gender. Find the probability that a randomly chosen student is male given the student’s bedroom is painted with dark colors. a. 30/ b. 30/ c. 22/ d. 30/ Question 17 A health club is interested in knowing how many times a typical member uses the club in a week. They decide to ask every tenth customer on a specified day to complete a short survey,
including information about how many times they have visited the club in the past week. What kind of a sampling design is this? a. Cluster b. Stratified c. Simple random d. Systematic Question 18 Which measure of the center of data would a clothing store use when placing orders for the typical middle customer? a. Mean b. Median c. Mode d. IQR Question 19 One hundred eighteen students were asked what type of color their bedrooms were painted: light colors, dark colors, or vibrant colors. The results were tabulated according to gender. Find the probability that a randomly chosen student is male or has a bedroom painted with light colors. a. 10/ b. 68/ c. 48/ d. 10/ Question 20 A community college offers classes six days a week: Monday through Saturday. Maria conducted a study of the students in her classes to determine how many days per week the students who are in her classes come to campus for classes. In each of her five classes she randomly selected 10 students and asked them how many days they come to campus for classes. Each of her classes are the same size. The results of her survey are summarized in the table below: How many students come to campus for classes four days a week? a. 49 b. 25 c. 30 d. 13
To find the expected number of cars parked crookedly, we multiply the total number of cars by the probability of a car being parked crookedly. Expected number of cars = Total number of cars * Probability of a car being parked crookedly So, Expected number of cars = 22 * 0.375 = 8. So, the answer is a. 8.. Question 3 We are interested in the number of times a teenager must be reminded to do his or her chores each week. A survey of 40 mothers was conducted. The table below shows the results of the survey. Find the probability that a teenager is reminded two times. a. 8 b. 8/ c. 6/ d. 2
Question 4 If X ~ Exp (0.8), then P (x < μ) = — a. 0. b. 0. c. 0. d. cannot be determined Properties of the exponential distribution (parameter given for the distribution) Exponential Distribution
- If X ∼Exp( λ ), then the probability density function (PDF) for an exponential distribution is given by:
- The mean (or expected value, μ ) of the exponential distribution is given by the formula: λ =0.8. Determine the Probability P ( X < μ ) To find P ( X < μ ) , we need to compute the cumulative distribution function (CDF) for the exponential distribution. The CDF for an exponential distribution is given by: F ( x )=1− e − λx Substitute μ into the CDF Formula We want P ( X < μ ) = P ( X <1.25): P ( X <1.25) = F (1.25) = 1− e −0.8⋅1. Calculate e −0.8⋅1. 0.8⋅1.25=1. e −1.0≈0. Compute F (1.25) P ( X <1.25) = 1− e −1^ ≈ 1−0.3679 = 0. Question 5 The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the z-score for a patient who takes 10 days to recover? a. 1. b. 0. c. 2. d. 7. Ussing the formula for the z-score: where: X is the value we're interested in (10 days recovery time), μ is the mean (5.3 days), σ is the standard deviation (2.1 days). Given Values:
- Standard Deviation σ Step 2: Calculate the Standard Error of the Mean (SEM) The standard error of the mean for a sample of size n is given by: Step 3: Finding the 90th Percentile of the Sample Mean The z-score for the 90th percentile in a standard normal distribution is approximately z = 1.2816. To find the 90th percentile of the sample average wait time, we use the z-score formula: X = μ + z * SEM Substituting the values: X = 37.5+ (1.2816 *2.165) Calculating z * SEM: 1.2816 * 2.165 ≈ 2. Now substituting that back into the equation: X ≈ 37.5+2.77 ≈ 40.27 Rounded to 40. Question 8 Richard’s Furniture Company delivers furniture from 10 a.m. to 2 p.m. continuously and uniformly. We are interested in how long (in hours) past the 10 a.m. start time that individuals wait for their delivery. The average wait time is: a. one hour b. two hours c. two and a half hours d. four hours Question 9 Identify the given random variable as being discrete or continuous. What is the braking time of a car? a. Discrete b. Continuous Time = Continous Question 10 The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the probability of spending more than two days in recovery?
a. 0. b. 0. c. 0. d. 0. Given Data: Mean μ = 5.3 days Standard deviation σ = 2.1 days P(X > 2) Calculate the z-score X = 2 Find the Probability P(X > 2) is equivalent to ( 1 - P(X<= 2)). Look up the z-score ( -1.5714 ) in the standard normal (Z) table, or use a calculator (or software). P(Z <= -1.5714) ≈ 0. Calculate the complement P(X > 2) = 1 - P(X <= 2) ≈ 1 - 0.0582 = 0. d. 0. Question 11 The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. What's the approximate probability that the average price for 16 gas stations is more than $4.69? a. almost zero b. 0. c. 0. d. unknown Find the approximate probability that the average cost of gasoline for 16 randomly chosen gas stations in the Bay Area exceeds $4.69, given that the population mean is $4.59 and the population standard deviation is $0.10. The distribution of the population is unknown, but we can use the Central Limit Theorem (CLT) since we’re dealing with a sample mean. Apply the Central Limit Theorem The CLT states that for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normal, regardless of the population distribution, provided the population has a finite mean and variance. Here, the sample size is 16, which is reasonably large (typically, n≥30 ) is considered sufficient, but (n = 16) is often acceptable in practice, especially with a symmetric or unimodal underlying distribution). We’re given:
- Population mean, μ = 4.
- Population standard deviation, σ = 0.
In the scenario described, Maria selected students from each of her five classes, with 10 students randomly selected from each class. This means she is not only using a convenience sampling method (due to the fact that she is selecting students from her own classes), but she is also ensuring that every class is represented in the sample. Analyzing the Sampling Techniques
- Convenience Sampling: Maria is using convenience sampling because she is selecting students who are readily available from her classes.
- Additional Technique Used:
- Since she is taking a specified number of students (10) from each of her five classes, she is ensuring that her sample includes students from each class equally. This approach allows her to analyze the responses in a way that represents different groups (the individual classes).
- This indicates a stratification based on the class groups she has. The sampling technique that is combined with convenience sampling is Stratified Sampling , as she is intentionally selecting students from each of her classes, effectively treating each class as a "stratum." Question 13 What kind of data is the amount of money spend in a grocery store? a. Qualitative b. Quantitative-continuous c. Quantitative-discrete IDK The correct answer should be Continuous, but the quiz says the correct answer is Discrete. An amount of money spent is Continuous, not Discrete. I think there's a mistake in this answer. Question 14 The lifetime of a computer circuit board is normally distributed with a mean of 2,500 hours and a standard deviation of 60 hours. What is the probability that a randomly chosen board will last at most 2,560 hours? a. 0. b. 0. c. 0. d. 0. Given the values X = 2, μ = 2,500 hours σ = 60 hours z = (2560 - 2500) / 60 = 60 / 60 = 1 Finding the value of Z in the Standard normal distribution table z = 1
P (Z ≤ 1) ≈ 0.
Question 15 A study finds that the mean amount spent in a grocery store per visit by customers in a sample is $12.84. This is an example of a: a. Population b. Sample c. Parameter d. Statistic e. Variable A statistic is a characteristic or measure obtained by using the data values from a sample. Question 16 In one town, the number of burglaries in a week has a Poisson distribution with a mean of 1.9. Find the probability that in a randomly selected week the number of burglaries is at least three. a. 0. b. 0. c. 0. d. 0. e. 0. Ussing the Poisson distribution formula. The Poisson distribution formula is given by: P(X = k) = (e^(-λ) * λ^k) / k! P(X = k) is the probability of getting exactly k events e is the base of the natural logarithm (approximately 2.71828) λ is the mean number of events in the given time period k is the number of events In this case, the mean number of burglaries in a week is 1.9. We want to find the probability that the number of burglaries is at least three, so we need to calculate the probability of getting three or more burglaries. P(X ≥ 3) = 1 - P(X < 3) To calculate P(X < 3), we can sum the probabilities of getting 0, 1, and 2 burglaries. P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) Using the Poisson distribution formula, we can calculate each of these probabilities: P(X = 0) = (e^(-1.9) * 1.9^0) / 0! = e^(-1.9) ≈ 0. P(X = 1) = (e^(-1.9) * 1.9^1) / 1! = 1.9 * e^(-1.9) ≈ 0. P(X = 2) = (e^(-1.9) * 1.9^2) / 2! = (1.9^2 * e^(-1.9)) / 2 ≈ 0. Now, we can calculate P(X < 3): P(X < 3) = 0.1496 + 0.2835 + 0.2698 ≈ 0. Finally, we can calculate P(X ≥ 3) by subtracting P(X < 3) from 1: P(X ≥ 3) = 1 - P(X < 3) ≈ 1 - 0.7030 ≈ 0.
distribution. This means that if we take enough samples, the average of these samples (the sample mean) should be close to the population mean. In this case, we have a sample size of 100 riders, which is large enough for the CLT to apply. Therefore, we would expect the sample mean to be close to the population mean of 37. minutes. Given the above, if the sample average wait time for 100 riders was less than 30 minutes, it would be surprising. This is because, according to the properties of the uniform distribution and the Central Limit Theorem, we would expect the sample mean to be close to the population mean of 37.5 minutes. So, the answer is: YES Question 20 The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days. What is the median recovery time? a. 2. b. 5. c. 7. d. 2. The mean is the average of a set of data, while the standard deviation is a measure of how spread out the data is. In a normal distribution, the median is the same as the mean.
