##### Question

In: Statistics and Probability

# The length, XX, of a fish from a particular mountain lake in Idaho is normally distributed...

The length, XX, of a fish from a particular mountain lake in Idaho is normally distributed with μ=9.1μ=9.1 inches and σ=2σ=2 inches.

(a) Is XX a discrete or continuous random variable? (Type: DISCRETE or CONTINUOUS)

(b) Write the event ''a fish chosen has a length equal to 6.1 inches'' in terms of XX:  .

(c) Find the probability of this event:

d) Find the probability that the length of a chosen fish was greater than 10.6 inches:  .

(e) Find the probability that the length of a chosen fish was between 6.1 and 10.6 inches:

## Solutions

##### Expert Solution

Solution:

R.V. X follows normal distribution with

$url=https://private.codecogs.com/gif.latex?%5Cmu&t=1571337672&ymreqid=50870fb5-0ab3-84d7-1c61-cb006c01b100&sig=IReUvGWR_gwc3HyBA5PaEA--~C$ = 9.1

$url=https://private.codecogs.com/gif.latex?%5Csigma&t=1571337672&ymreqid=50870fb5-0ab3-84d7-1c61-cb006c01b100&sig=9R.Hdbs5eeZZZxj1SUatuw--~C$ = 2

a) Normal is a continuous type distribution.

So , answer is CONTINUOUS

b) event ''a fish chosen has a length equal to 6.1 inches''

In terms of X , event can be written as " X = 6.1"

c) P(X = 6.1) = 0

Because , for a continuous random variable , the probability at a particular point is considered as zero.

d)P( greater than 10.6 inches)

P(X > 10.6) = P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=kvg_MFGfmDVnnUbLmMnBWQ--~C$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=bTRZkVySMHrJhzIGJIKFRQ--~C$ >  (10.6 - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=kvg_MFGfmDVnnUbLmMnBWQ--~C$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=bTRZkVySMHrJhzIGJIKFRQ--~C$]

= P[Z > (10.6 - 9.1)/2]

= P[Z > 0.75]

= 1 - P[Z < 0.75]

=    1 - 0.7734 ( use z table)

=    0.2266

e)P(between 6.1 and 10.6 inches)

= P(6.1 < x< 10.6)

= P(X < 10.6) - P(X < 6.1)

=  P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=kvg_MFGfmDVnnUbLmMnBWQ--~C$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=bTRZkVySMHrJhzIGJIKFRQ--~C$ <  (10.6 - 9.1)/2] -   P[(X - $url=https://latex.codecogs.com/gif.latex?%5Cmu&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=kvg_MFGfmDVnnUbLmMnBWQ--~C$ )/$url=https://latex.codecogs.com/gif.latex?%5Csigma&t=1571335224&ymreqid=50870fb5-0ab3-84d7-1c61-cb003a01b100&sig=bTRZkVySMHrJhzIGJIKFRQ--~C$ <  (6.1 - 9.1)/2]

= P[Z < 0.75] - P[Z < -1.50]

= 0.7734 - 0.0668 ..Use z table

= 0.7066

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