Printable View
The ages of ships using a port is normally distributed with a mean age of 12 years and standard deviation of 5 years.
State the probability that a ship will be more than 15 years old?
I've hit this question several times and got 0.2257
I think i have missed a step.
If someone could please confirm or correct it and show simply how they did it so I can learn from my mistake that would be great.
You need to convert to the standard normal distribution, then look on your table of standard normal distribution values.Quote:
Originally Posted by student1234
Your right but I don't know if the value I converted is correct. i'm hoping that someone could show me how to do it from the start. I have no confidence in this
That won't happen. What you'll need to do is post everything you have done, then if there are any errors we can help put them right.Quote:
Originally Posted by student1234
Mean= 12
S.D= 5
X= 15
Z= X-Mean = (15 - 12) /5
S.D
Z score= 0.6
Probability Conversion= 0.2257
Thats what I did to get the Z-score I'm sure I've missed something though
You are correct that the z score is 0.6, so you are trying to evaluate Pr(Z > 0.6). From the symmetry of the normal curve, this is equal to Pr(Z < -0.6).Quote:
Originally Posted by student1234
Can you read this value off your table?
It says 0.2451 Whats the next step
You are looking at the wrong part of the table... Pr(Z < -0.6) = 0.2743
This is the answer.
Thank you
So is there any difference in the procedure if the question says for example 'age of ships less than 10 year'? (basically less than the mean)
No difference at all. You would evaluate the z value, then work out Pr(Z < z).
Cool Thanks