3.29) The shell of the land snail Limocolaria martensiana has two possible color

Question

3.29)   The shell of the land snail Limocolaria martensiana has two possible colors forms: streaked and pallid. In a certain population of these snails, 60% of the individuals have streaked shells. Suppose that a random sample of 10 snails is to be chosen from this population. Define Y as the number of streaked-shelled snails in the sample of 10 snails and use the Y variable within probability notation to find the probability that the percentage of streaked-shelled snails in the sample will be those requested in problems 29a-c.

A) P(Y = 50%)

B)  P(Y = 60%)

C) P(Y = 70%)

D) P(Y = 0.50)

E) P(Y = 0.60)

F) P(Y = 0.70)

G) P(Y = 5)

H) P(Y = 6)

I) P(Y = 7)

=Answer?

=Answer?

3.30)    Consider taking a sample of size 10 from the snail population in Exercise 3.29. Find the following:

Use 2nd decimal accuracy.

  Use 2nd decimal accuracy.

A) P(Y = 50%)

B)  P(Y = 60%)

C) P(Y = 70%)

D) P(Y = 0.50)

E) P(Y = 0.60)

F) P(Y = 0.70)

G) P(Y = 5)

H) P(Y = 6)

I) P(Y = 7)

Explanation / Answer

3.29a)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.6      
x = the number of successes =    5      
          
Thus, the probability is          
          
P (    5   ) =    0.200658125 = 0.2007 [ANSWER]

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3.29B)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.6      
x = the number of successes =    6      
          
Thus, the probability is          
          
P (    6   ) =    0.250822656 = 0.2508 [answer]

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3.29c)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    10      
p = the probability of a success =    0.6      
x = the number of successes =    7      
          
Thus, the probability is          
          
P (    7   ) =    0.214990848 = 0.2150 [ANSWER]

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