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Scilla [17]
2 years ago
5

8. The heights of female students at North Shore High School are normally distributed with a mean of 175 cm and a standard devia

tion of 6 cm. About how many students in a random sample of 1000 female students could we expect to have heights less than 167.5 cm
Mathematics
1 answer:
gulaghasi [49]2 years ago
8 0

The number of students with heights less than 163 cm should be expected is 12.

According to the statement

The mean of height is 175 cm

and  the standard deviation of height is 6 cm.

We use normal distribution here with formula

Z= X - μ /σ

Here X is 167.5 and μ is 175 and σ is 6 cm.

Substitute the values in it then

Z = 167.5 - 175 / 6

Z = -1.25

-1.25 have a p value 0.012

Out of 1000 students:

0.012 x 1000 = 12.

So, The number of students with heights less than 163 cm should be expected is 12.

Learn more about DISTRIBUTION here brainly.com/question/1094036

#SPJ4

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It is 1/2 because if you add a extra zero to each fraction, it will look like this:

10/20

80/15

The 10/20 is bigger because 20 is bigger than 15.
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The measure of an angle is five times the measure of its complementary angle. What is the measure of each angle?
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Step-by-step explanation:

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  2. So the angle you want is 5x  
  3. Then [x + 5x] = 180 because they are supplementary.  
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The ratio of girls to boys in a bicycling club was 2:3 they were 27 boys how many total members were there in the club
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The National Highway Traffic Safety Administration reported the percentage of traffic accidents occurring each day of the week.
Zigmanuir [339]

Answer:

Hence,

a)

Test statistic X^{2} = 9.269.  

p value = 0.148

b)

Sunday=  15.23%, Monday= 12.62%, Tuesday= 13.10%, Wednesday=11.67%,

Thursday= 13.10%, Friday= 16.19%, Saturday= 18.09%.

Step-by-step explanation:

a)

                    relative observed    Expected       Chi square

Category   frequency(p)      O_{i}    Ei=total*p    (O-E)^{2} R^{2}_{i} =(O_{i} -E_{i} )^{2} /E_{i}

     1           0.142857143      64.0         60.00    16.00          0.267

    2           0.142857143      53.0       60.00     49.00         0.817

    3           0.142857143      55.0         60.00     25.00        0.417

    4           0.142857143      49.0       60.00     121.00        2.017

    5           0.142857143      55.0         60.00     25.00        0.417

    6           0.142857143      68.0         60.00     64.00        1.067

    7           0.142857143      76.0         60.00    256.00        4.267

  Total         1.000      420        420             556       9.269

Test statistic X^{2} = 9.269.  

p value = 0.148 from excel: chidist(9.269,6)  

b)

Percentage %= Frequency of traffic occur on category x 100 / Total number of frequency given.

Category          Percentage%

Sunday                   15.23%

Monday                    12.62%

Tuesday                   13.10%

Wednesday           11.67%

Thursday           13.10%

Friday                   16.19%

Saturday                   18.09%

5 0
3 years ago
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