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vfiekz [6]
3 years ago
6

What is the eighth letter in this question not Counting spaces? W is the first H is the second etc

Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0
The eighth letter is ‘H’
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Can you add the whole question I think it may be 6/4
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One environmental group did a study of recycling habits in a California community. It found that 75% of the aluminum cans sold i
AysviL [449]

Answer:

a

  P( \^ p  >  0.775 ) =  0.12798

b

 P( 0.6718 < p  <  0.775 ) =0.87183

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.75

Considering question a  

     The sample size is  n  =  387

Generally the standard deviation of this sampling distribution is  

         \sigma  = \sqrt{ \frac{p(1 - p)}{ n } }    

=>      \sigma  = \sqrt{ \frac{0.75(1 - 0.75)}{ 387 } }    

=>      \sigma  = 0.022    

The sample proportion of cans that are recycled is

                 \^ p =  \frac{ 300}{387 }

=>              \^ p =  0.775

Generally the probability that 300 or more will be recycled is mathematically represented as

         P( \^ p  >  0.775 ) =  P( \frac{\^ p  -  p }{ \sigma }  >  \frac{0.775 - 0.75 }{ 0.022} )

\frac{\^ p  - p }{\sigma }  =  Z (The  \ standardized \  value\  of  \ \^ p  )

       P( \^ p  >  0.775 ) =  P( Z >  1.136  )

From the z table  the area under the normal curve to the left corresponding to  1.591   is

      P( Z >  1.136)  = 0.12798

=>    P( \^ p  >  0.775 ) =  0.12798

Considering question b

Generally the lower limit of  sample proportion of cans that are recycled is

                 \^ p_1 =  \frac{ 260 }{387 }

=>              \^ p_1  =  0.6718

Generally the upper limit of  sample proportion of cans that are recycled is

                 \^ p_2 =  \frac{ 300}{387 }

=>              \^ p_2  =  0.775

Generally probability that between 260 and 300 will be recycled is mathematically represented as

           P( 0.6718 < p  <  0.775 ) =  P( \frac{0.6718 - 0.75 }{ 0.022}<  \frac{\^ p  -  p }{ \sigma }

=>      P( 0.6718 < p  <  0.775 ) =  P( -3.55 <  Z < 1.136 )

=>        P( 0.6718 < p  <  0.775 ) = P(Z <  1.136 ) -  P( Z <  -3.55 )

From the z table  the area under the normal curve to the left corresponding to  1.136 and  -3.55  is

       P( Z <  -3.55 ) = 0.00019262

and

       P(Z <  1.136 )  = 0.87202

So

       P( 0.6718 < p  <  0.775 ) =  0.87202-  0.00019262

=>   P( 0.6718 < p  <  0.775 ) =0.87183

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