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kkurt [141]
3 years ago
8

Alligators captured in Florida are found to have a mean length of 2 meters and a standard deviation of 0.35 meters. The lengths

of alligators are believed to be approximately normally distributed. What is the approximate length of an alligator at the 67th percentile of alligator lengths?
a. 2.01 meters.
b. 2.44 meters.
c. 2.21 meters.
d. 2.15 meters.
e. 2.09 meters.
Mathematics
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

The correct option is (d). 2.15 meters.

Step-by-step explanation:

According to the given scenario, the approximate length of an alligator at the 67th percentile is as follows:

The length of an alligator is X \sim N(2,0.35) .

Let us assume the P_{67} be the 67th percentile

P(X

\frac{P_{67 - 2 }}{0.35} = 0.44\\\\P_{67} = 2.154

Hence, the correct option is (d). 2.15 meters

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Answer:

a) \simeq 0.3012   b) \simeq 0.0494 c) \simeq 0.2438

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Rate of collision,

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So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

          P(X =x) = \frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}


                                                           for x ∈ N ∪ {0}

                       =  0 otherwise --------------------------------------(1)

here, \lambda = 0.3 collision / month

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         P(X = 0) = \frac{e^{-0.3}\times {\0.3}^{0}}{0!}


                      \simeq 0.7408182207 ------------------------------------(2)

Now, since we are calculating  this for 4 months,

so, P(No collision in 4 month period)

     =0.7408182207^{4}

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Putting X =1 in (1) we get,

           P(X =1) = \frac{e^{-0.3}\times {\0.3}^{1}}{1!}


                      \simeq 0.2222454662 ------------------------------------(4)

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P(2 collisions in 2 month period)

                =0.2222454662^{2}

                \simeq 0.0494 -----------------------------------------(5)

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= P( X = 1/6]  (which is to be estimated)

=\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

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P(1  collision in 6 month period)

  =  0.6543894283^{6}

   \simeq 0.0785267444 ------------------------------------------------(7)

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   \simeq 0.1652988882 ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

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