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evablogger [386]
1 year ago
14

A sample has a mean of M = 90 and a standard devia-tion of s 20.=a. Find the z-score for each of the following X values.X = 95X

= 80X = 98X = 88X = 105X = 76
Mathematics
1 answer:
sladkih [1.3K]1 year ago
5 0

Answer:

\begin{gathered} X=95,z=0.25 \\ X=80,z=-0.5 \\ X=98,z=0.4 \\ X=88,z=-0.1 \\ X=105,z=0.75 \\ X=76,z=-0.7 \end{gathered}

Explanation:

Given a sample with the following:

• Mean,M = 90

,

• Standard deviation, s = 20

To find the z-score for each of the given X values, we use the formula below:

\begin{equation*} z-score=\frac{X-\mu}{\sigma}\text{ where }\begin{cases}{X=Raw\;Score} \\ {\mu=mean} \\ {\sigma=Standard\;Deviation}\end{cases} \end{equation*}

The z-scores are calculated below:

\begin{gathered} \text{When X=95, }z=\frac{95-90}{20}=\frac{5}{20}=0.25 \\ \text{When X=80, }z=\frac{80-90}{20}=\frac{-10}{20}=-0.5 \\ \text{When X=98, }z=\frac{98-90}{20}=\frac{8}{20}=0.4 \end{gathered}\begin{gathered} \text{When X=88,}z=\frac{88-90}{20}=\frac{-2}{20}=-0.1 \\ \text{When X=105, }z=\frac{105-90}{20}=\frac{15}{20}=0.75 \\ \text{When X=76, }z=\frac{76-90}{20}=\frac{-14}{20}=-0.7 \end{gathered}

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Vedmedyk [2.9K]

Given:

The distance between the two buildings on a map = 14 cm

The scale is 1:35000.

To find:

The actual distance in km.

Solution:

The scale is 1:35000.

It means 1 cm on map = 35000 cm in actual.

Using this conversion, we get

14 cm on map = 14\times 35000 cm in actual.

                       = 490000 cm in actual.

                       = 4.9\times 1000o0 cm in actual.

                       = 4.9 km in actual.          [1\text{ km}=100000\text{ cm}]

Therefore, the actual distance between two buildings is 4.9 km.        

5 0
3 years ago
3. How many silver cars is the dealer expected to sell, on average, out of five cars? Interpret this expected value in context.
ivann1987 [24]

Answer:

E(X)= 0*0.27068 +1*0.40426+ 2*0.24151+3*0.07214+4*0.01077+5*0.00064=1.14998

So then the expected number of silver cars to sell for this case in the next five car sales is approximately 1.15 for the 23% of the total manufacturing cars produced.

Step-by-step explanation:

Assuming this problem :"23% of the cars a certain automaker manufactures are silver. Below is the probability distribution for the number of  silver cars sold by a car dealer in the next five car sales."

The distirbution is given by:

Num. Silver cars     0      |          1   |     2     |       3    |       4     |       5     |

Probability        0.27068|0.40426|0.24151|0.07214|0.01077|0.00064|

And assuming the following questions

What is the probability of selling at most three silver cars? Interpret this probability in context

For this case we just need to find this:

P(X\leq 3) = P(X=0) +P(X=1) +P(X=2)+P(X=3) = 0.27068+0.40426+0.24151+0.07214=0.98859

So then the probability of selling at most 3 silver cars in the next five car salesfrom the 23% of the total manufactured cars is 0.98859.

What is the probability of selling between and including one and four silver cars? Interpret this probability in  context.

For this case we just need to find this:

P(1 \leq X\leq 4) = P(X=1) +P(X=2) +P(X=3)+P(X=4) = 0.40426+0.24151+0.07214+0.01077=0.72868

So then the probability of selling between 1 and 4 silver cars in the next five car sales from the 23% of the total manufactured cars is 0.72868  in the next five car sales.

How many silver cars is the dealer expected to sell, on average, out of five cars? Interpret this expected value in context

For this case we can find the expected value on this way:

E(X)= 0*0.27068 +1*0.40426+ 2*0.24151+3*0.07214+4*0.01077+5*0.00064=1.14998

So then the expected number of silver cars to sell for this case in the next five car sales is approximately 1.15 for the 23% of the total manufacturing cars produced.

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3 years ago
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810 total feet because 135 times 6 is 810
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