Solve the problem that involves Normal distribution

[chwala]

Homework Statement

See attached

Relevant Equations

stats

My interest is on part (c),
My take,

$Z=\dfrac{160−200}{60}=−0.666666$

$Pr(−0.66666)=0.3546$

$⇒\dfrac{x_1-200}{60}=1.05$

$x_1=63+200=263$

Yes, i am aware that they want the answer to $5$ significant figures...i just wanted to check the alternative method...
Appreciate your insight...

 

[Maarten Havinga]

What do you mean with the probability of -2/3? The density of the distribution there?

 

[Mark44]

Homework Statement:: See attached
Relevant Equations:: stats

View attachment 302276View attachment 302277My interest is on part (c),
My take,

$Z=\dfrac{160−200}{60}=−0.666666$

$Pr(−0.66666)=0.3546$

One approach is to calculate P(X > 160). Your work should include the probabilities that some random variable is greater than or less than some number. Pr(-.66666) is meaningless as a probability, a pointed hinted at by the previous poster.

To get you started, calculate Pr(X > 160) = Pr($\frac{X - 160}{60} > \frac{160 - 200}{60}$) = Pr(Z > -0.66667)##. That should give you a number that is larger than 0.6. You should then be able to find the Z score (and hence the X value) that gives you the correct probability, using a table of z-scores.

$⇒\dfrac{x_1-200}{60}=1.05$

$x_1=63+200=263$

Yes, i am aware that they want the answer to $5$ significant figures...i just wanted to check the alternative method...
Appreciate your insight...

 

[chwala]

One approach is to calculate P(X > 160). Your work should include the probabilities that some random variable is greater than or less than some number. Pr(-.66666) is meaningless as a probability, a pointed hinted at by the previous poster.

To get you started, calculate Pr(X > 160) = Pr($\frac{X - 160}{60} > \frac{160 - 200}{60}$) = Pr(Z > -0.66667)##. That should give you a number that is larger than 0.6. You should then be able to find the Z score (and hence the X value) that gives you the correct probability, using a table of z-scores.

My point was that when $z =-0.66666$ Then the area subtended between the two x points would be equal to $0.3546$

 

[chwala]

What do you mean with the probability of -2/3? The density of the distribution there?

Yes I meant the density...I shouldn't have used term probability...I hope that makes sense...otherwise, I need to check again...and I should have used inequalities instead of equal sign...Will amend that later.

 

[Mark44]

My point was that when $z =-0.66666$

Then that is incorrect since Pr(Z = -0.66666) = 0. For a continuous probability, any probability statement should involve a random variable in an inequality.

 

[Maarten Havinga]

It's OK, making mistakes is how we learn ;)

 

[chwala]

It's OK, making mistakes is how we learn ;)

Yes, in my approach i first found the area bound by $160$ and $X.$ I found that to be; $0.2454$ then i took $0.6-0.2454=0.3542$ ...

Homework Statement:: See attached
Relevant Equations:: stats

View attachment 302276View attachment 302277My interest is on part (c),
My take,

$Z=\dfrac{160−200}{60}=−0.666666$

$Pr(−0.66666)=0.2454$

$⇒\dfrac{x_1-200}{60}=1.05$

$x_1=63+200=263$

Yes, i am aware that they want the answer to $5$ significant figures...i just wanted to check the alternative method...
Appreciate your insight...

Homework Statement:: See attached
Relevant Equations:: stats

View attachment 302276View attachment 302277My interest is on part (c),
My take,

$Z=\dfrac{160−200}{60}=−0.666666$

$Pr(−0.66666)=0.3546$

$⇒\dfrac{x_1-200}{60}=1.05$

$x_1=63+200=263$

Yes, i am aware that they want the answer to $5$ significant figures...i just wanted to check the alternative method...
Appreciate your insight...

I made a mistake ...find the corrected version here;

My interest is on part (c),
My take,

$Z=\dfrac{160−200}{60}=−0.666666$

$Pr(−0.66666)=0.2454$

$0.6-0.2454=0.3542$

$0.3542= Z_{1.05}$ Therefore

$⇒\dfrac{x_1-200}{60}=1.05$

$x_1=63+200=263$

Yes, i am aware that they want the answer to $5$ significant figures...i just wanted to check the alternative method...
Appreciate your insight...