Tassone K Midterm
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Yavapai College *
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167
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Mathematics
Date
Apr 3, 2024
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docx
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Uploaded by ProfessorJellyfishPerson3801
Kaitlyn Tassone
MAT 167 Fall 2023
WA 1 Midterm
Activity #1:
A: Does improvisation in jazz music enhance creativity more than other genres of music
?
B: The participants in this study are the 12 jazz improvisers, the 12 classical musicians, and 12 musicians. The sample strategy would possibly be convenient sampling because those selected participated willingly. C: This study is an experiment since the researchers controlled the chord progressions type that participant listened and responded to. D: The response variables are the participants’ liking of the different cord progressions as well as
their responses to them. So, this would be a qualitative variable.
E: The results show that the jazz musicians preferred the unexpected riffs whereas the improvisers had faster and stronger neural responses to the unusual music. The classical musicians hadn’t figured out the surprising music. F: The researchers concluded that increase in creativity can come from training to be receptive to the unexpected area of expertise like jazz improvisation. Activity #2:
A: This histogram is skewed right and shows that the highest frequency was between 50-64.9 which means that majority of the states have identity theft rates between 50 and 80 per 100,000 of the population. B: The mean of identity thefts per 100,000 population is 64.45 and the standard deviation is 23.82. Therefore, the number of identity thefts per 100,000 population in each state deviates from the mean by about 23.82. The median of 61.8 best represents the center as it has less outliers or extreme values.
C: The boxplot has a minimum of 24.6, first quartile of 47.1, median of 61.8, third quartile of 77.4, and a maximum of 153.4. The boxplot shows a spread of the quartiles as 30.3 and an outlier of 153.4 as the maximum.
Activity #3:
A: In a random phenomenon the two possible terms used to label the outcomes would be “Success” and Failure”. Some examples would be flipping a coin, playing in a basket basketball game, rolling a die, and scratching a lottery ticket. B: The probability of success, p, for free throws is the same for each trial because each free throw is an independent event, and the outcome of one trial does not affect the outcome of another.
C: The probability of inheriting sickle cell disease for a child with two parents who are carriers is 25% (0.25). Each child’s genetic makeup is determined independently of their siblings therefore the probability would be the same for each child in the family.
D: The four conditions needed for a binomial distribution is there is a fixed number of trials, each trial is independent, there are only two possible outcomes for each trial, and the probability of success is the same for each trial. E: I: The binomial setting has a fixed number of trials (n) however, this number of trials (rolls) is not fixed as the rolls continue until a six appears. II: This situation is within the binomial setting as there are a fixed number of trials (n= 10), each trial is independent, and each trial results in one of two outcomes. The probability of success (p= 0.6).
III: This situation is within the binomial setting as there are a fixed number of trials (accidents in a week), each trial is independent, and each trial results in one of two outcomes (involving alcohol or not). However, we can’t determine n and p because we don’t know the total number of accidents or the proportion that involves alcohol.
I: This situation is not within the binomial setting because the probability of success (drawing a red card) changes with each draw as cards are not replaced.
Extra Credit: The probability that all cards are red in a standard deck with 26 red cards and 52 in total would be (26/52) * (25/51) * (24/50) * (23/49) * (22/48) = 0.000181.
Activity #4:
A: Using the empirical rule, the following intervals of time in seconds it takes for Wendy’s drive through would be, 68% (within one SD of mean): 113.5-163.5 seconds, 95% (within two SDs of mean): 88.5-188.5 seconds, 99.5% (within three SDs of mean): 63.5 to 213.5 seconds. So, 68% of the times are between 138.5-25 (113.5) and 138.5+25 (163.5) seconds, 95% are between 138.5-2*25 (88.5) and 138.5+2*25 (188.5) seconds, and 99.5% are between 138.5-3*(63.5) and 138.5+3*25 (213.5) seconds.
B: The z-score that corresponds to a drive-through time of 145 seconds is 0.26. So,
it would be calculated as (145-138.5)/25 = 0.26
C: The probability that a car will spend more than 145 seconds is the area to the right of 0.26 on the standard normal distribution. It would be 1 – 0.6026 = 0.3974.
D: The z-scores for 120 and 180 are -0.74 and 1.66. The probability that a car will spend between 120 and 180 seconds is the area between these z-scores which would be 0.8790 – 0.0604 = 0.8186.
E: one minute is more than three standard deviations below the mean, therefore it would be unusual. F: The z-score for the 75
th
percentile is 0.67. So, the corresponding drive-through time would be 151.3 seconds.
Activity #5:
A
: The mean would be the average height of all 20 children and the standard deviation would measure the dispersion of the heights from the mean. The
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sampling
distribution (the means of all samples of size 4 from the population), however, would tend to follow a normal distribution (according to the Central Limit Theorem) with the same mean as the population but a smaller standard deviation. The distributions are alike in that they both have the same mean, but they differ in their shape and standard deviation.
B: A control chart is a statistical tool used in quality control processes to determine if a manufacturing or business process is in a state of control. It shows the variation in a measurement over time and compares it to defined upper and lower control limits. If the process stays within these limits, it is considered "in control". If it goes outside these limits, it is considered "out of control" and corrective action is needed.
C: The population distribution of the duration of calls for one month might be skewed or have outliers, depending on the nature of the calls. However, the sampling distribution of the means of the duration of calls would tend to become more normally distributed as the sample size gets larger, according to the Central Limit Theorem.
D: The Central Limit Theorem states that the average of your sample means will be the population mean, and as you take more samples, especially large ones, your graph of the sample means will look more like a normal distribution curve.
Activity #6:
A: The probability that a male is randomly selected, and weight is greater than 175
would be calculated as z = (175 - 180.5) / 32.4 = -0.17. Looking up this z-score in a standard normal distribution table, we find that the probability is approximately 0.57.
B: The mean of the sampling distribution is equal to the population mean, which is 180.5 pounds. The standard deviation of the sampling distribution is equal to the population standard deviation divided by the square root of the sample size, which is 32.4 / √150 = 2.64 pounds.
C:
To find the probability that the mean weight of the 150 passengers is greater than 175 pounds, we calculate the given values, so we get z = (175 - 180.5) /
(32.4 / √150) = -13.13. Looking up this z-score in a standard normal distribution table, we find that the probability is approximately 1.
D: The pilot should not be worried about exceeding the weight limit because the probability that the mean weight of the passengers is greater than 175 pounds is almost certain, but this is still below the estimated weight of 180.5 pounds per passenger. Therefore, the total weight of the passengers is likely to be within the weight limit.