normal distributions (fish in lake, etc)

can someone please help me with some of these quesions?

If X is uniformly distributed over a range of values 8 to 21, then what is:

a. the value of the probability density function f(x)?

0.0769 ? is it correct?

b. mean and standard deviation of X?

mean is 14.5 and SD is 3.75 is it correct?

c. pr(10 ≤ X ≤ 17)?

0.538 is it correct?

d. pr(X > 22)?

e. pr(X ≥ 7)?

A statistics lecturer and a mathematics lecturer went fishing at a lake during a break at a conference in England. Beforehand they agreed that the one who caught the more impressive fish would have their dinner paid for by the other. The statistics lecturer caught a 40 cm long perch, while the mathematics lecturer caught a 42 cm bream.

From long term studies of the fish populations in the lake, it is known that the length of perch is approximately normally distributed with mean of 29.6 cm and standard deviation of 9.5 cm. The length of bream is approximately normally distributed with mean of 38.4 cm and standard deviation of 4.2 cm.

The next day the mathematics lecturer went fishing again and caught 1 bream and 1 perch. What is the probability that the bream is longer than the perch? What assumption was made to do this calculation?

Answers (a) - (c) are correct.

For (d) and (e), draw a graph of the uniform density, which is a rectangle. Does the rectangle have any area above x = 22? How much of the rectangle's area is above x = 7?

For the word problem, consider the variable Z equal to the difference between the length of the bream and the length of the perch. Pr(Z > 0) is the desired probability. Z is a normal variable whose mean is the difference in the means and whose variance is the sum of the variances under a certain assumption. What's the assumption?

hi

cheers for the reply.

so

for d) is it probability of 0 ?
and
e) how much is the area above 7?

theres the values 8-22 so probability of 1.

is probability of 1 correct?


for the word problem, could u do it or part of it ?

the difference between E(b-p)=8.8 and VAR(b-p)=107.89 --->SD=10.39(2dp)

then i done

P[X>(8.8-29.6)/10.39]
and got
P(Z>-2)

from negative Z table i get
P(Z>-2)=0.023

is this correct?

Hi, you're welcome.

(d) and (e) are correct.

For the word problem, the mean and standard deviation are correct. The probability should be

\(\displaystyle \L \begin{array}{ll}
Pr(Z\ >\ 0) &= Pr(\ (Z - 8.8)/10.39\ >\ (0 - 8.8)/10.39\ ) \\
&= 1 - \Phi(-8.8/10.39) \\
&\approx .80 , \\
\end{array}\)

where \(\displaystyle \L \Phi\) gives the table value from the standard cumulative normal distribution function.

thanks, i will look more into that