Question

see attachment

Answer 1

is it a no?

Answer 2

P(Y <= (174.88 - mean)/std) ) where Y is the normal random distribution

Answer 3

so X is not 3 (apples?)

Answer 4

ok i guess not! it cant be anyway.. so what is the 3 apples for?

Answer 5

Three apples throws a wrench into the problem...What is the distribution of the mean weight of 3 apples?

Answer 6

I would expect the mean to be the same, but the variance to be different

Answer 7

I say no also, because this would not resemble the normal distribution for only 3 apples, selected at random -- because of the variance dependence on the number of samples

Answer 8

yeah thats what i thought as well. i ll submit in a bit and i will let you know if the answer was correct.. :) thanks @ybarrap

Answer 9

ok

Answer 10

oh i think i know how to solve this.. u have to use original std deviation x \[\sqrt{3}\] apples.. then u just have to take 174.88-157/std deviation x \[\sqrt{3}\

Answer 11

i was thinking that as well, but wan't 100%

Answer 12

yeah, so the working i gave above should be correct right? cos u said std dev shudnt be the same

Answer 13

i got the answer but its not in the choices given..

Answer 14

that would be my best guess. Only problem I have is that std/sqrt(n) is the standard deviation from the expected mean. Is that what we want? Usually when applying this technique, we want the standard deviation from the sample's mean. Which sounds like what we want.

Answer 15

omg, its std dev/square root.. lol im so blur.. i did multiplication..

Answer 16

So you looked up P(Y < 1.63) ?

Answer 17

yup! got the same thing.. which is 0.9484

Answer 18

looks good

Answer 19

aite thanks a bunch! wud update u with the answer as im trying to solve other questions as well before submitting :D

Answer 20

yes

Answer 21

how about this question @ybarrap

Answer 22

The upper limit of the confidence interval is
\[157=\bar{x}+z\frac{19}{\sqrt{3}}\]
Substituting 174.88 for the sample mean gives
\[157=174.88+z\frac{19}{\sqrt{3}}\]
Rearranging we get
\[z=\frac{174.88-157}{\frac{19}{\sqrt{3}}}=1.63\]
Reference to a standard normal distribution table using the z-score 1.63 shows the required probability is 0.9484

Answer 23

oh i figured.. its x= 7500.69 grams/50

Answer 24

then just do the normal equation thing like before

Answer 25

Same type of problem, but instead of \(\mu\) it would be \(n*\mu\) and variance would be \( n^{2}*Var(x)\), because now we are talking about a sum of n normally-distributed random variables.

Answer 26

so it shud be 7500.69/50= 150.0138, (150.0138-157)/(19/sqr root 50)=-2.6
so probability is 1-0.047=-2.6

Answer 27

P( Sum > 7500.69) = P( (Sum - n*mean)/(n*19) > (7500.69 - n*mean)/(n*19) )

Answer 28

how did u get -2.56.. is my working above correct tho?

Answer 29

@ybarrap did u get z=-2.6 as well?

Answer 30

@bambimonster Sorry, my bad. -2.6 is correct.

Answer 31

okay thank you :)

Answer 32

P( Sum > 7500.69) = P( (Sum - n*mean)/(sqrt(n)*19) > (7500.69 - n*mean)/(sqrt(n)*19) )
=P(Y > -2.6), where Y is normally distributed

Answer 33

The probability that the weight of the sample is less than 7500.69 grams is 0.0047, corresponding to a z-score of -2.6. Therefore the probability that the weight of the sample is greater than 7500.69 grams is 1.0000 - 0.0047 = ?

Answer 34

my answer is wrong... or isit -2.59.. cos i calculated again and it was -2.59 so it shud be 1-0.0048=0.9952?

Answer 35

That's what I found, but I using that variance of a sum of n normal variables is n*Var(x)

Answer 36

approximate, definitely. But the means converge in distribution to the normal distribution as the number of sample gets larger by the central limit theorem

Answer 37

so shud i add that part to it?

Answer 38

No, that's essentially what you already have.

Answer 39

so i shudnt add that part cos i already explained?

Answer 40

yea, you said as n get larger it approaches normal distribution. That's what CLT says.

Answer 41

oh okay.. sorry im very blur today.. :(

Answer 42

sound clear-minded to me

Answer 43

nah its like i ask questions that i already know the answer to it.. not those that idk.. and then i m making silly mistakes .. smh.. anyway thanks i got it all correct :)

Answer 44

nice

Answer 45

help @ybarrap

Answer 46

The CLT applies to any distribution's mean, regardless of what that distribution is. I think that you can use the same procedure as before, taking into account the sampling rate (i.e. dividing by sqrt(n)). You CAN calculate it, but you would not expect that the accuracy to be very high with such low sample. Part B is the same as we had before. I say that I would not want to use the process for the 3 samples, but would not have a problem using it for the 35 samples. Sorry for the ambiguity.

Answer 47

I hope this helps. I need to bug out for the night, but I will be back on tomorrow evening.

Answer 48

i used the same procedure.. but i didnt get it.. so yeah .. okay not a problem.. good night!

Answer 49

lots of other great help around here