Question

Probability question regarding normal population distribution. Screenshot attached

Answer 1

I do not understand what I'm doing wrong for the last two parts.

\[P(T_0=8)\]Applying standardization for \(T_0\):\[Z=\frac{ T_0-n \mu }{ \sigma \sqrt{n} }\]
Then finding probability of Z in the standard normal curve tables that we're given.

But clearly they're not accepting it. What am I doing wrong here?

Answer 2

How can you ever have P(x>=7) less than P(x=8) ?

Answer 3

Notice that P(x>=7) = P(x=7) + P(x=8)

Answer 4

I may be wrong, but if you have more attempts, could you try entering below answers :

0.0625
0.1125

Answer 5

@ganeshie8 Yeah, those are correct! What did you do?

Answer 6

Wow!
I've got those using your method in part a..

Answer 7

Interesting...

Answer 8

P(x=8) = 0.5^4
= 0.0625

P(x>=7) = P(x=7) + P(x=8)
= 4*0.5^3*0.1 + 0.0625
= 0.1125

Answer 9

Why 0.5^4?

Answer 10

Think of an \(4\) digit number :
\[abcd\]

and the digits can be taken from the set \(\{0,1,2\}\)

Answer 11

To have sum of digits equal to 8, each digit must equal 2.
That's the only way we can ever get a sum equal to 8, yes ?

Answer 12

I apologize, I don't think I really grasp what you're explaining. .
There's x1 and x2 and both are equal.. T0=x1+x2.. So I don't understand your approach

Answer 13

\(T_0 = x_1+x_2+x_3+x_4\)

and \(x_i\in \{0,1,2\}\)

how many ways are there for \(T_0\) to equal \(8\) ?

Answer 14

Ah I see, each x would have to be 2.. *facepalm*

Answer 15

haha right
second part follows from same reasoning..

Answer 16

So how did you get the exponent, then?

Answer 17

\(T_0 = x_1+x_2+x_3+x_4\)

and \(x_i\in \{0,1,2\}\)

For \(T_0\) to equal \(8\), we must have \(x_1=x_2=x_3=x_4=2\)

We are given \(P(x_i=2) \) is \(0.5\)

since the variables \(x_i\) are independent :
\(P(T_0=8) = 0.5*0.5*0.5*0.5=0.5^4\)

Answer 18

Yeah that makes sense! Thanks!

Answer 19

np :)