In all of William Shakespeare\'s works, he used 884,647 different words. Of thes

Question

In all of William Shakespeare's works, he used 884,647 different words. Of these 14,376 appeared only once. In 1985, a 429-word poem was discovered that may have been written by Shakespeare. To keep the probability calculations simple, assume that the choices between a new word and one from the list of 884,647 are independent for each of the 429 words. Approximate a porbability that a new word will not be on the list, by the relative frequencey of words used once.

a) Find the expected number of new words in the poem

b) Use the normal approximation to the binomial to determine the probability of finding 12 or more new words in the poem. Use the continuity correction.

c) use the normal approximation to the binomial to determine the probability of finding 2 or fewer new words in the poem. Use continuity correction.

d) Use the normal approximation to the binomial to determine the probability of finding more than 2 but less than 12 new words in the poem.

Explanation / Answer

a)

The probability of a new word is

p = 14376/884647 = 0.01625055

Hence, as n = 429,

E(x) = n p = 429*0.01625055 = 6.97148595 [ANSWER]

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b)

We first get the z score for the critical value:          
          
x = critical value =    11.5      
u = mean = np =    6.97148595      
          
s = standard deviation = sqrt(np(1-p)) =    2.618815662      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    1.729222151      
          
Thus, the right tailed area is          
          
P(z >   1.729222151   ) =    0.041884672 [ANSWER]

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c)

We first get the z score for the critical value:          
          
x = critical value =    2.5      
u = mean = np =    6.97148595      
          
s = standard deviation = sqrt(np(1-p)) =    2.618815662      
          
Thus, the corresponding z score is          
          
z = (x-u)/s =    -1.707445856      
          
Thus, the left tailed area is          
          
P(z <   -1.707445856   ) =    0.043869607 [ANSWER]

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d)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    2.5      
x2 = upper bound =    11.5      
u = mean = np =    6.97148595      
          
s = standard deviation = sqrt(np(1-p)) =    2.618815662      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -1.707445856      
z2 = upper z score = (x2 - u) / s =    1.729222151      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.043869607      
P(z < z2) =    0.958115328      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.914245722   [ANSWER]  

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