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Ganezh [65]
2 years ago
12

Mrs William has a prized box with different colored tickets. Each ticket results in different type of prize.

Mathematics
1 answer:
Lera25 [3.4K]2 years ago
6 0

\text{ Answer:} \text{ Probability that he chooses a yellow ticket and then a purple ticket }= \frac{3}{25}

Step-by-step explanation:

Since we have given that

Number of green tickets =3

Number of yellow tickets = 4

Number of purple tickets = 3

Total number of tickets = 3+4+3=10

Now,

As we know that Pedro select a ticket and replace it and then select another ticket , total number of outcomes will never change it will be 10 only for each case.

Probability that he chooses a yellow ticket and then a purple ticket is given by

\frac{4}{10}\times \frac{3}{10}\\\\=\frac{12}{100}\\\\=\frac{3}{25}

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Sedaia [141]

Complete Question

Answer:

a

  SE  = 0.66}

b

-3.29 <  \mu_1 - \mu_2 <  -0.70  

Step-by-step explanation:

From the question we are told that

  The sample size is  n  = 60

   The first sample mean is  \= x _1  =  8

    The second sample mean is   \= x _2  =  10

    The first variance is  v_1 =  0.25

    The first variance is  v_2 =  0.55

Given that  the confidence level is 95% then the level of significance is 5% =  0.05

Generally from the normal distribution table the critical value  of  \frac{\alpha }{2} is  

   Z_{\frac{\alpha }{2} } =  1.96

Generally the first standard deviation is  

     \sigma_1 =  \sqrt{v_1}

=>   \sigma_1 =  \sqrt{0.25}

=>   \sigma_1 =  0.5

Generally the second standard deviation is

     \sigma_2 =  \sqrt{v_2}

=>   \sigma_2 =  \sqrt{0.55}

=>   \sigma_2 =  0.742    

Generally the first standard error is

     SE_1  =  \frac{\sigma_1}{\sqrt{n} }

      SE_1  =  \frac{0.5}{\sqrt{60} }

     SE_1  =  0.06

Generally the second standard error is

     SE_2  =  \frac{\sigma_2}{\sqrt{n} }

      SE_2  =  \frac{0.742}{\sqrt{60} }

     SE_2  =  0.09

Generally the standard error of the difference between their mean scores is mathematically represented as    

      SE  =  \sqrt{SE_1^2 + SE_2^2 }

=>     SE  =  \sqrt{0.06^2 +0.09^2 }

=>     SE  = 0.66}

Generally 95% confidence interval is mathematically represented as  

      (\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } *  SE) <  \mu_1 - \mu_2 <  (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } *  SE)

=> (8 -10) -(1.96 *  0.66) <  \mu_1 - \mu_2 <  (8-10) +(Z_{\frac{\alpha }{2} } *  0.66)  

=>  -3.29 <  \mu_1 - \mu_2 <  -0.70  

 

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