find the z - score. round results to the nearest hundredth.\n15) the mean of a set of data is 108.06 and its…

find the z - score. round results to the nearest hundredth.\n15) the mean of a set of data is 108.06 and its standard deviation is 115.45. find the z - score for a value of 369.67.\na) 3.84\nb) 2.98\nc) 3.31\nd) 3.61\n15) \nfind the range for the given sample data.\n16) fred, a local mechanic, recorded the price of an oil and filter change at twelve competing service stations. the prices (in dollars) are shown below.\n32.99 24.95 26.95 28.95\n18.95 28.99 30.95 22.95\n24.95 26.95 29.95 28.95\na) $14.04\nb) $10.05\nc) $12.00\nd) $12.99\n16) \nfind the midrange for the given sample data.\n17) a meteorologist records the number of clear days in a given year in each of 21 different u.s. cities. the results are shown below.\n72 143 52 84 100 98 101\n120 99 121 86 60 59 71\n125 130 121 104 74 83 55 169\na) 98 days\nb) 112 days\nc) 117 days\nd) 110.5 days\n17) \nfind the mean of the data summarized in the given frequency distribution.\n18) the heights of a group of professional basketball players are summarized in the frequency distribution below. find the mean height. round your answer to one decimal place.\nheight (in.) | frequency\n70 - 71 | 1\n72 - 73 | 8\n74 - 75 | 13\n76 - 77 | 8\n78 - 79 | 15\n80 - 81 | 6\n82 - 83 | 3\na) 75.5 in.\nb) 78.1 in.\nc) 76.6 in.\nd) 75.3 in.\n18) \nfind the standard deviation for the probability distribution. round to the nearest hundredth.\n19) the random variable x is the number of houses sold by a realtor in a single month at the sendsoms real estate office. its probability distribution is as follows.\nhouses sold (x) | probability p(x)\n0 | 0.24\n1 | 0.01\n2 | 0.12\n3 | 0.16\n4 | 0.01\n5 | 0.14\n6 | 0.11\n7 | 0.21\na) σ = 2.62\nb) σ = 4.45\nc) σ = 6.86\nd) σ = 2.25\n19)

find the z - score. round results to the nearest hundredth.\n15) the mean of a set of data is 108.06 and its standard deviation is 115.45. find the z - score for a value of 369.67.\na) 3.84\nb) 2.98\nc) 3.31\nd) 3.61\n15) \nfind the range for the given sample data.\n16) fred, a local mechanic, recorded the price of an oil and filter change at twelve competing service stations. the prices (in dollars) are shown below.\n32.99 24.95 26.95 28.95\n18.95 28.99 30.95 22.95\n24.95 26.95 29.95 28.95\na) $14.04\nb) $10.05\nc) $12.00\nd) $12.99\n16) \nfind the midrange for the given sample data.\n17) a meteorologist records the number of clear days in a given year in each of 21 different u.s. cities. the results are shown below.\n72 143 52 84 100 98 101\n120 99 121 86 60 59 71\n125 130 121 104 74 83 55 169\na) 98 days\nb) 112 days\nc) 117 days\nd) 110.5 days\n17) \nfind the mean of the data summarized in the given frequency distribution.\n18) the heights of a group of professional basketball players are summarized in the frequency distribution below. find the mean height. round your answer to one decimal place.\nheight (in.) | frequency\n70 - 71 | 1\n72 - 73 | 8\n74 - 75 | 13\n76 - 77 | 8\n78 - 79 | 15\n80 - 81 | 6\n82 - 83 | 3\na) 75.5 in.\nb) 78.1 in.\nc) 76.6 in.\nd) 75.3 in.\n18) \nfind the standard deviation for the probability distribution. round to the nearest hundredth.\n19) the random variable x is the number of houses sold by a realtor in a single month at the sendsoms real estate office. its probability distribution is as follows.\nhouses sold (x) | probability p(x)\n0 | 0.24\n1 | 0.01\n2 | 0.12\n3 | 0.16\n4 | 0.01\n5 | 0.14\n6 | 0.11\n7 | 0.21\na) σ = 2.62\nb) σ = 4.45\nc) σ = 6.86\nd) σ = 2.25\n19)

Answer

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x - \mu}{\sigma}$, where $x$ is the data - value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 108.06$, $\sigma=115.45$, and $x = 489.67$. $z=\frac{489.67 - 108.06}{115.45}$

Step2: Calculate the numerator

$489.67−108.06 = 381.61$

Step3: Calculate the z - score

$z=\frac{381.61}{115.45}\approx3.31$

Answer:

C. 3.31

Explanation:

Step1: Recall range formula

The range of a data - set is given by $R=\text{Max}-\text{Min}$. For the oil and filter change prices: $\text{Max}=32.99$, $\text{Min}=18.95$. $R = 32.99−18.95$

Step2: Calculate the range

$R=14.04$

Answer:

A. $14.04$

Explanation:

Step1: Recall mid - range formula

The mid - range of a data - set is given by $\text{Mid - Range}=\frac{\text{Max}+\text{Min}}{2}$. For the number of clear days data: $\text{Max}=143$, $\text{Min}=52$. $\text{Mid - Range}=\frac{143 + 52}{2}$

Step2: Calculate the mid - range

$\text{Mid - Range}=\frac{195}{2}=97.5\approx98$ (rounded to the nearest whole number)

Answer:

A. 98 days

Explanation:

Step1: Calculate the mid - points of each class

For the height intervals:

Height (in.) Frequency ($f$) Mid - point ($x$) $f\times x$
70 - 71 1 70.5 70.5
72 - 73 8 72.5 580
74 - 75 13 74.5 968.5
76 - 77 8 76.5 612
78 - 79 15 78.5 1177.5
80 - 81 6 80.5 483
82 - 83 3 82.5 247.5

Step2: Calculate the sum of $f\times x$ and the sum of $f$

$\sum f=1 + 8+13 + 8+15+6+3=54$ $\sum(f\times x)=70.5 + 580+968.5+612+1177.5+483+247.5 = 4139$

Step3: Calculate the mean

$\bar{x}=\frac{\sum(f\times x)}{\sum f}=\frac{4139}{54}\approx76.6$

Answer:

C. 76.6 in.

Explanation:

Step1: Recall the formula for the mean of a probability distribution

$\mu=\sum(x\times P(x))$ $\mu=(0\times0.24)+(1\times0.01)+(2\times0.12)+(3\times0.16)+(4\times0.01)+(5\times0.14)+(6\times0.11)+(7\times0.21)$ $=0 + 0.01+0.24 + 0.48+0.04+0.7+0.66+1.47=3.6$

Step2: Recall the formula for the variance of a probability distribution

$\sigma^{2}=\sum((x-\mu)^{2}\times P(x))$ $(0 - 3.6)^{2}\times0.24+(1 - 3.6)^{2}\times0.01+(2 - 3.6)^{2}\times0.12+(3 - 3.6)^{2}\times0.16+(4 - 3.6)^{2}\times0.01+(5 - 3.6)^{2}\times0.14+(6 - 3.6)^{2}\times0.11+(7 - 3.6)^{2}\times0.21$ $=( - 3.6)^{2}\times0.24+( - 2.6)^{2}\times0.01+( - 1.6)^{2}\times0.12+( - 0.6)^{2}\times0.16+(0.4)^{2}\times0.01+(1.4)^{2}\times0.14+(2.4)^{2}\times0.11+(3.4)^{2}\times0.21$ $=12.96\times0.24 + 6.76\times0.01+2.56\times0.12+0.36\times0.16+0.16\times0.01+1.96\times0.14+5.76\times0.11+11.56\times0.21$ $=3.1104+0.0676+0.3072+0.0576+0.0016+0.2744+0.6336+2.4276 = 6.88$

Step3: Calculate the standard deviation

$\sigma=\sqrt{\sigma^{2}}=\sqrt{6.88}\approx2.62$

Answer:

A. $\sigma = 2.62$