Question

a commercial prawn fisherman recorded the number of prawns he took from each trap. The average number of prawns per trap was 230 with a standard deviation of 22. What number of prawns per trap woud you expect in the interval symmetrical about the mean where 80% of the numbers would be found?

Ps. I don't have a graphing calculator and I need to answer the question for an assignment I can email in today. Thus please explain and give the calculations :)

Answer 1

You first need the z-values covering the required interval. Reference to the standard normal distribution shows these to be z = 1.281, z = -1.281.
\[z=\frac{X-\mu}{\sigma}\]
Mu is the parent mean and sigma is the standard deviation. Can you find X ?

Answer 2

Sorry. The negative value of z should be used. Can you calculate with this value?

Answer 3

uhhhh i don't know.... but that was wrong

Answer 4

You calculated correctly with the positive value of z. Just change the sign of z and calculate again.

Answer 5

okay

Answer 6

got it

Answer 7

What did you get for X ?

Answer 8

233.16

Answer 9

\[-1.281=\frac{X-230}{22}\]
\[X=230-(1.281\times 22)=\]

Answer 10

What do you get for X now ?

Answer 11

201.818

Answer 12

Correct! Rounded up it is 202. Therefore there are 202 prawns in the interval symmetrical about the mean where 80% of the numbers are found.

Answer 13

thank you so much!

Answer 14

You're welcome :)