Question

Weights of female cats of a certain breed are normally distributed with mean 4.3 kg and standard deviation 0.6 kg.

What proportion of female cats have weights between 3.7 and 4.4 kg?

How heavy is a female cat whose weight is on the 80th percentile?

A female cat is chosen at random. What is the probability that she weighs more than 4.5 kg?

Answer 1

solution :

Given that ,

mean = = 4.3

standard deviation = = 0.6

P(3.7< x <4.4 ) = P[(3.7-4.3) / 0.6< (x - ) / < (4.4-4.3) / 0.6)]

= P( -1< Z <0.17 )

= P(Z < 0.17) - P(Z <-1 )

Using z table   

= 0.5675-0.1587

= 0.4088

2.

How heavy is a female cat whose weight is on the 80th percentile?

solution

Using standard normal table,

P(Z < z) = 80%

= P(Z < z) = 0.80

= P(Z < 0.84) = 0.80

z = 0.84

Using z-score formula  

x= z * +

x= 0.84*0.6+4.3

x= 4.804

3.A female cat is chosen at random. What is the probability that she weighs more than 4.5 kg?

solution:

P(x >4.5 ) = 1 - P(x<4.5 )

= 1 - P[(x -) / < (4.5-4.3) /0.6 ]

= 1 - P(z <0.33 )

Using z table

= 1 -0.6293

= 0.3707

probability= 0.3707