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Karolina [17]
3 years ago
10

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2

and 5.1.
A)about 68%
B)about 34%
C)about 16%
D)about 13.5%
Mathematics
1 answer:
Yuliya22 [10]3 years ago
7 0

A set of data has a normal distribution with a mean of 5.1 and a standard deviation of 0.9. Find the percent of data between 4.2 and 5.1.

Answer: The correct option is B) about 34%

Proof:

We have to find P(4.2

To find P(4.2, we need to use z score formula:

When x = 4.2, we have:

z = \frac{x-\mu}{\sigma}

          =\frac{4.2-5.1}{0.9}=\frac{-0.9}{0.9}=-1

When x = 5.1, we have:

z = \frac{x-\mu}{\sigma}

          =\frac{5.1-5.1}{0.9}=0

Therefore, we have to find P(-1

Using the standard normal table, we have:

P(-1= P(z

                               =0.50-0.1587

                               =0.3413 or 34.13%

                               = 34% approximately

Therefore, the percent of data between 4.2 and 5.1 is about 34%

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48 ft^2

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What is the inequality 4/5__1/2
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Answer:

Step-by-step explanation:

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● 4/5 = 0.8

0.8 is greater than 0.5.

So:

● 0.8 > 0.5

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3 years ago
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A random sample of 77 fields of corn has a mean yield of 26.226.2 bushels per acre and standard deviation of 2.322.32 bushels pe
PSYCHO15rus [73]

Answer:

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

Step-by-step explanation:

n = 77

mean u = 26,226.2  bushels per acre

standard deviation s = 2,322.32

let E = true mean

let A = test statistic

Find 95% Confidence Interval

so

let  A =  (u - E) *  (\sqrt{n}  / s)   be the test statistic

we want      P( average_l <  A  < average_u )  = 95%

look for  lower 2.5%  and the upper 97.5%  Because I think this is a 2-tail test

average_l =  -1.96  which corresponds to the 2.5%

average_u = 1.96

P(  -1.96  <  A  <  1.96)  =  95%

P(  -1.96  <  (u - E) *  (\sqrt{n}  / s)  <  1.96)  =  95%

Solve for the true mean E ok

E   <   u + 1.96* (s  / \sqrt{n})

from  -1.96  <  (u - E) *  (\sqrt{n}  / s)

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  1.96*( 2,322.32 / \sqrt{77} )

E < 26,226.2 +  518.7197348105429466

upper bound is 26,744.9197

or

u - 1.96* (s  / \sqrt{n})  < E

26,226.2 -  518.7197348105429466  < E

25,707.48026519  < E

lower bound is 25,707.48026519

Therefore the  95% confidence interval is (25,707.480 < E < 26,744.920)

7 0
3 years ago
Recently, six single-family homes in San Luis Obispo County in California sold at the following prices (in $1,000s): 560, 468, 6
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The interval for mean sale price is $582

According to statement

The given rates are 560, 468, 685, 534, 658, 593

Now,

Sample mean = X / n

Put the values in it and then

(560+ 468+ 685+ 534+ 658+ 593) / 6 = 582.1667

and use this formula S = sqrt [ X2 - n 2 / (n-1) ]

and put the values in these and

S = sqrt [ X2 - n 2 / (n-1) ] = 95.97179

So, The interval for mean sale price is $582

Learn more about SAMPLE MEAN here

brainly.com/question/16794295

#SPJ4

6 0
1 year ago
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