10) which of the following statements is true? events m and e are \na) mutually exclusive\nc) complement\nb)…

10) which of the following statements is true? events m and e are \na) mutually exclusive\nc) complement\nb) not exclusive\nd) independent\n11) what is the probability that the employee has only one degree, p(e∩m^c) or (m∩e^c) = _?\na) 15/34\nb) 17/34\nc) 19/34\nd) 21/34\nevaluate the expression.\n12) which of the following is equal to _np_4/_nc_4 (for n > 4)\na) 6\nb) 24\nc) 120\nd) 720\nsolve the problem.\n13) you have 6 letters consisting of one i, two as and three ms. if the letters are randomly arranged in order, what is the probability that the arrangement spells the word \mamami\? hint: the number of the partition is needed!\na) 1/720\nb) 1/360\nc) 1/60\nd) 1/120\nfind the mean of the data summarized in the given frequency distribution.\n14) the manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one monday. the frequency distribution below summarizes the results. find the mean waiting time. round your answer to one decimal place.\nwaiting time (minutes) | numebr of customers | xmid | fxmid\nll ~ ul | f | |\n0 ~ 8 | 14 | 4 | 56\n9 ~ 17 | 20 | 13 | 260\n18 ~ 26 | 6 | 22 | 132\na) 9.6 min\nb) 10.5 min\nc) 11.2 min\nd) 12.9 min\nsolve the problem.\n15) if {x_1,x_2,......,x_n} with average of 8, range of 30, risk of 7, then ∑x^2 - n(x̅)^2/(n - 1)= _\na) 64\nb) 49\nc) 900\nd) 15\n16) in problems 16 through 18, given the whisker - and - box plot of a data set of the weights (in pounds) of 30 newborn babies as {x_1,x_2,x_3, ...,x_30} and ∑x = 225.0 and ∑x^2 = 1948.5. find the second quartile q_2 = _\na) 7.7\nb) 1.3\nc) 7.0\nd) 6.4\n17) find sample mean x̅ = _\na) 7.4\nb) 7.5\nc) 7.0\nd) 7.2

10) which of the following statements is true? events m and e are \na) mutually exclusive\nc) complement\nb) not exclusive\nd) independent\n11) what is the probability that the employee has only one degree, p(e∩m^c) or (m∩e^c) = _?\na) 15/34\nb) 17/34\nc) 19/34\nd) 21/34\nevaluate the expression.\n12) which of the following is equal to _np_4/_nc_4 (for n > 4)\na) 6\nb) 24\nc) 120\nd) 720\nsolve the problem.\n13) you have 6 letters consisting of one i, two as and three ms. if the letters are randomly arranged in order, what is the probability that the arrangement spells the word \mamami\? hint: the number of the partition is needed!\na) 1/720\nb) 1/360\nc) 1/60\nd) 1/120\nfind the mean of the data summarized in the given frequency distribution.\n14) the manager of a bank recorded the amount of time each customer spent waiting in line during peak business hours one monday. the frequency distribution below summarizes the results. find the mean waiting time. round your answer to one decimal place.\nwaiting time (minutes) | numebr of customers | xmid | fxmid\nll ~ ul | f | |\n0 ~ 8 | 14 | 4 | 56\n9 ~ 17 | 20 | 13 | 260\n18 ~ 26 | 6 | 22 | 132\na) 9.6 min\nb) 10.5 min\nc) 11.2 min\nd) 12.9 min\nsolve the problem.\n15) if {x_1,x_2,......,x_n} with average of 8, range of 30, risk of 7, then ∑x^2 - n(x̅)^2/(n - 1)= _\na) 64\nb) 49\nc) 900\nd) 15\n16) in problems 16 through 18, given the whisker - and - box plot of a data set of the weights (in pounds) of 30 newborn babies as {x_1,x_2,x_3, ...,x_30} and ∑x = 225.0 and ∑x^2 = 1948.5. find the second quartile q_2 = _\na) 7.7\nb) 1.3\nc) 7.0\nd) 6.4\n17) find sample mean x̅ = _\na) 7.4\nb) 7.5\nc) 7.0\nd) 7.2

Answer

Explanation:

Step1: Recall permutation and combination formulas

The formula for permutation ${n}P{r}=\frac{n!}{(n - r)!}$ and combination ${n}C{r}=\frac{n!}{r!(n - r)!}$. So, $\frac{{n}P{4}}{{n}C{4}}=\frac{\frac{n!}{(n - 4)!}}{\frac{n!}{4!(n - 4)!}}$.

Step2: Simplify the expression

$\frac{\frac{n!}{(n - 4)!}}{\frac{n!}{4!(n - 4)!}}=\frac{n!}{(n - 4)!}\times\frac{4!(n - 4)!}{n!}=4!$.

Step3: Calculate factorial value

Since $4!=4\times3\times2\times1 = 24$.

Answer:

  1. B. 24

Step1: Calculate total number of arrangements of 6 letters

The formula for permutations of multi - set with $n$ objects where there are $n_1$ of one kind, $n_2$ of another kind, $\cdots$, $n_k$ of the $k$th kind is $\frac{n!}{n_1!n_2!\cdots n_k!}$. Here $n = 6$ (total letters), $n_1=1$ (number of I's), $n_2 = 2$ (number of A's) and $n_3=3$ (number of M's). So the total number of arrangements is $\frac{6!}{1!2!3!}=\frac{720}{1\times2\times6}=60$.

Step2: Calculate probability

There is only 1 way to spell "MAMAMI". So the probability $P=\frac{1}{60}$.

Answer:

  1. C. $\frac{1}{60}$

Step1: Recall mean formula for grouped data

The mean $\bar{x}=\frac{\sum fX_{mid}}{\sum f}$, where $f$ is the frequency and $X_{mid}$ is the mid - point of the class interval. $\sum f=14 + 20+6=40$ and $\sum fX_{mid}=56 + 260+132=448$.

Step2: Calculate the mean

$\bar{x}=\frac{448}{40}=11.2$.

Answer:

  1. C. 11.2 min

Step1: Recall the formula for sample variance

The formula for sample variance $s^{2}=\frac{\sum x^{2}-n(\bar{x})^{2}}{n - 1}$. The risk given here is likely the standard deviation $s$. Since $s = 7$, and $s^{2}=\frac{\sum x^{2}-n(\bar{x})^{2}}{n - 1}$, then $\frac{\sum x^{2}-n(\bar{x})^{2}}{n - 1}=s^{2}=49$.

Answer:

  1. B. 49

Step1: Recall the definition of the second quartile in a box - and - whisker plot

The second quartile $Q_2$ is the median of the data set. In a box - and - whisker plot, the line inside the box represents the median. From the box - and - whisker plot, the value of the line inside the box is 7.0.

Answer:

  1. C. 7.0

Step1: Recall sample mean formula

The sample mean $\bar{X}=\frac{\sum X}{n}$. Given $\sum X = 225.0$ and $n = 30$.

Step2: Calculate the sample mean

$\bar{X}=\frac{225.0}{30}=7.5$.

Answer:

  1. B. 7.5