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Lera25 [3.4K]
3 years ago
10

Every day a kindergarten class chooses randomly one of the 50 state flags to hang on the wall, without regard to previous choice

s, We are interested in the flags that are chosen on Monday, Tuesday and Wednesday of next week.
a) Describe a sample space \Omega and a probability measure P to model this experiment.

b) What is the probability that the class hangs Wisconsin's flag on Monday, Michigan's flag on Tuesday, and California's flag on Wednesday.?

c) What is the probability that Wisconsin's flag will be hung at least two of the three days?
Mathematics
1 answer:
Svet_ta [14]3 years ago
7 0

Answer:

a.)  P(x = X) = \frac{1}{50}

b.) \frac{1}{50} \times\frac{1}{50} \times\frac{1}{50}  = \frac{1}{125000}

c.) 0.00118

Step-by-step explanation:

The sample space Ω = flags of all 50 states

a.) Any one of the flags is randomly chosen. Therefore we can write the    

   probability measure as P(x = X) = \frac{1}{50} , for all the elements of the sample

   space, that is for all x ∈ Ω.

b.) the probability that the class hangs Wisconsin's flag on Monday,

   Michigan's flag on Tuesday, and California's flag on Wednesday

 = \frac{1}{50} \times\frac{1}{50} \times\frac{1}{50}  = \frac{1}{125000}

c.) the probability that Wisconsin's flag will be hung at least two of the three days

= Probability that Wisconsin's flag will be hung on two days + Probability that Wisconsin's flag will be hung on three days

= P(x = 2) + P(x = 3)

= (\binom{3}{2}\times \frac{1}{50} \times \frac{1}{50}\times \frac{49}{50}) + (\binom{3}{3}\times \frac{1}{50} \times \frac{1}{50}\times \frac{1}{50})\\

= \frac{147}{125000} + \frac{1}{125000}

= \frac{148}{125000}

= 0.00118

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

Yes the researcher can  conclude that the supplement has a significant effect on cognitive skill

b

d =  0.5778

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Step-by-step explanation:

From the question we are told that

  The sample size is  n=16

    The sample mean is  M  =  50.2

    The standard deviation is \sigma  = 9

    The  population mean is  \mu =  45

     The level of significance is  \alpha =  0.05

The null hypothesis is H_o  \mu =45

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Generally the test statistics is mathematically represented as

          z =  \frac{M -  \mu}{\frac{\sigma}{\sqrt{n} } }

=>       z =  \frac{50.2 - 45}{\frac{9}{\sqrt{16} } }

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Generally the p-value is mathematically represented as

     p-value  =  2 *  P(Z >  z )

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From the z-table  

      P(Z >  2.31 ) = 0.010444

=>     p-value  =  2 *  0.010444

=>     p-value  =  0.021

From the obtained values we see that p-value <  0.05

 Decision Rule

 Reject the null hypothesis

Conclusion

There is sufficient evidence to conclude that the supplement has a significant effect on the cognitive skill of elderly adults

Generally the Cohen's d for this study is mathematically represented as

    d =  \frac{M  -  \mu}{\sigma }

=>  d =  \frac{50.2 -45}{9 }

=>  d =  0.5778

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