mean and standard deviation (2 Viewers)

[starbuckscoffee]

the weight of cats is normally distributed. It is known that 10% of cats weigh more than 1.8kg, and 15% of cats weigh less than 1.35kg. Find the mean and the standard deviation of this distribution.

 

[Timothy.Siu]

i'm pretty sure thats not 2unit coz i hav no idea how to do that lol
but uhhh do u know wat is meant by normally distributed, it might help,

 

[Iruka]

the weight of cats is normally distributed. It is known that 10% of cats weigh more than 1.8kg, and 15% of cats weigh less than 1.35kg. Find the mean and the standard deviation of this distribution.

You have to convert to z-scores.

According to the table that I looked up, the 90th percentile corresponds to a z score of approx. 1.28, and the 25th percentile corresponds to a z-score of -0.67. So, if X is the mean and s the standard deviation,

(1.8-X)/ s = 1.28
(1.35-X)/s = -0.67

So we can obtain the two simultaneous equations

1.8-X = 1.28s (1)
1.35-X = -0.67s (2)

If we subtract eq (2) from eq (1), we have

0.45 = 1.95s, so s = 0.2308 (to 4 dp) and substituting this back into eq (1), we have X= 1.556

Timothy.Siu,
There's no stats in any of the calculus based courses yet, but if you do maths at uni you'll probably learn it in first year.

 

[PC]

Yeah, this is well outside the General Course.

 

[Iruka]

The General Maths course includes z-scores, but they don't use statistical tables to convert percentiles into z-scores (apart from the Empirical rule, that is). So I guess you could say it is outside the syllabus.