5 from unit 1, lesson 13 the mean exam score for the first group of twenty examinees applying for a security job is 35.1 with a standard deviation of 3.6. the mean exam score for the second group of twenty examinees is 34.1 with a standard deviation of 0.5. both distributions are close to symmetric in shape. a. use the mean and standard deviation to compare the scores of the two groups. b. the minimum score required to get an in - person interview is 33. which group do you think has more people get in - person interviews? 6 from unit 1, lesson 13 a group of pennies made in 2018 are weighed. the mean is approximately 2.5 grams with a standard deviation of 0.02 grams. interpret the mean and standard deviation in terms of the context. 7 from unit 1, lesson 12 these values represent the expected number of paintings a person will produce over the next 10 days. 0 0 0 1 1 1 2 2 3 5 a. what are the mean and standard deviation of the data? b. the artist is not pleased with these statistics. if the 5 is increased to a larger value, how does this affect the median, mean, and standard deviation?

Question

Accepted Answer

### 5a.
# Explanation:
## Step1: Analyze the means
The mean of the first - group is 35.1 and the mean of the second - group is 34.1. A higher mean in the first group indicates that, on average, the first - group of examinees scored higher on the exam.
## Step2: Analyze the standard deviations
The standard deviation of the first group is 3.6 and of the second group is 0.5. A smaller standard deviation in the second group means that the scores in the second group are more closely clustered around the mean, while the scores in the first group are more spread out.
# Answer:
On average, the first group of examinees scored higher (mean of 35.1 vs 34.1). The scores in the second group are more consistent (standard deviation of 0.5 vs 3.6).

### 5b.
# Explanation:
## Step1: Consider the position relative to the mean
The minimum score for an in - person interview is 33. The mean of the first group is 35.1 with a standard deviation of 3.6, and the mean of the second group is 34.1 with a standard deviation of 0.5. Since the first - group has a higher mean and a larger spread (standard deviation), more of its values are likely to be above 33.
# Answer:
The first group is likely to have more people get in - person interviews.

### 6.
# Explanation:
## Step1: Interpret the mean
The mean weight of 2.5 grams for the 2018 pennies represents the average weight of the pennies in the group. If we were to weigh all the 2018 pennies in the population (or a large sample), 2.5 grams is the central value around which the weights are centered.
## Step2: Interpret the standard deviation
The standard deviation of 0.02 grams indicates how much the weights of the pennies vary from the mean. A small standard deviation of 0.02 grams means that most of the 2018 pennies have weights that are very close to 2.5 grams, with only minor fluctuations.
# Answer:
The mean of 2.5 grams is the average weight of the 2018 pennies. The standard deviation of 0.02 grams shows that the weights of the pennies are closely clustered around the mean, with little variation.

### 7a.
# Explanation:
## Step1: Calculate the mean
The sum of the data values $0 + 0+0 + 1+1+1+2+2+3+5=15$. There are $n = 10$ data points. The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{15}{10}=1.5$.
## Step2: Calculate the variance
First, find the squared differences from the mean: $(0 - 1.5)^2=2.25$ (three times), $(1 - 1.5)^2 = 0.25$ (three times), $(2 - 1.5)^2=0.25$ (two times), $(3 - 1.5)^2 = 2.25$, $(5 - 1.5)^2=12.25$. The sum of the squared differences is $3\times2.25+3\times0.25 + 2\times0.25+2.25+12.25=6.75 + 0.75+0.5+2.25+12.25 = 22.5$. The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}=\frac{22.5}{9}=2.5$.
## Step3: Calculate the standard deviation
The standard deviation $s=\sqrt{s^{2}}=\sqrt{2.5}\approx1.58$.
# Answer:
The mean is 1.5 and the standard deviation is approximately 1.58.

### 7b.
# Explanation:
## Step1: Analyze the effect on the median
The data set has $n = 10$ values. The median is the average of the 5th and 6th ordered values, which are 1 and 1 currently, so the median is 1. Increasing the 5 to a larger value does not change the fact that the 5th and 6th ordered values remain 1 and 1, so the median remains 1.
## Step2: Analyze the effect on the mean
The mean $\bar{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}$. Increasing the 5 to a larger value will increase the sum of the data values $\sum_{i = 1}^{n}x_{i}$. Since $n$ is fixed at 10, the mean will increase.
## Step3: Analyze the effect on the standard deviation
The standard deviation measures the spread of the data. Increasing the 5 to a larger value will increase the squared differences from the mean for that data - point. Since the standard deviation is based on these squared differences, the standard deviation will increase.
# Answer:
The median remains the same, the mean increases, and the standard deviation increases.

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QUESTION IMAGE

Question

Explanation:

5a.

Step1: Analyze the means

Step2: Analyze the standard deviations

Step1: Consider the position relative to the mean

Step1: Interpret the mean

Step2: Interpret the standard deviation

Answer:

5b.