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marta [7]
1 year ago
15

A researcher is studying psychological factors in academic achievement among teenage girls. One variable she is particularly int

erested in is competitiveness. What information does a measure of variability for the variable competitiveness convey? Check all that apply. Do all teenage girls have the same amount of competitiveness? What is the average value for the variable competitiveness among teenage girls? How consistent are the values for competitiveness among teenage girls? What is the most common value for the variable competitiveness among teenage girls?
Mathematics
1 answer:
horrorfan [7]1 year ago
6 0

Answer:

The correct answer is:option A and B

Step-by-step explanation:

A researcher is studying and examining psychological factors in academic achievement amongst teenage boys. One of the variable he is particularly paying more attention is in determination.

The quantifying of variability for the variable determination suggests how stretched out are the values for determination amid teenage boys (do all boys that are still in their teenage years have the same determination) and how constant are the values of determination amid teenage boys.

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Answer:

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Step-by-step explanation:

From the question we are told that

  The  mean is  \mu =  7.5

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Generally the probability that the selected pool has a pH level that is not considered safe is mathematically represented as

       P(A) =  1 - P(7.2 \le X  \le 7.8 )

Here  

      P(7.2 < X  < 7.8 ) = P(\frac{ 7.2 - \mu }{\sigma } <  \frac{X - \mu }{ \sigma }

Generally \frac{X - \mu }{ \sigma } =  Z (The  \ standardized \  value  \  of  X )

So

 P(7.2 < X  < 7.8 ) = P(\frac{ 7.2 - 7.5 }{0.2 } < Z    

 P(7.2 < X  < 7.8 ) = P(-1.5 < Z  

=>  P(7.2 < X  < 7.8 ) = P(Z <  1.5) - P(  Z < - 1.5)

From the z-table the probability of  (Z <  -1.5) and   (  Z <1.5)  are

    P(Z <  1.5) =  0.93319

and  

    P(Z <  -1.5) =  0.066807

So

      P(7.2 < X  < 7.8 ) =0.93319 - 0.066807

       P(7.2 < X  < 7.8 ) =0.0866383

So

   P(A) =  1 - 0.0866383

=>  P(A) =  0.133617

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