Question

At two years of age, sardines inhabiting Japanese waters have a length distribution that is approximately normal with mean 20.20 cm and standard deviation of 0.65 cm.

a. how short are the shortest 15% of all these sardines?

b. in what range does the middle 70% of all sardines fall?

Answer 1

Solution :

mean = = 20.20 cm

standard deviation = = 0.65

Using standard normal table,

a ) P(Z < z) = 0.15%

P(Z < z) = 0.15

P(Z < - 1.036) = 0.15

z = - 1.04

Using z-score formula,

x = z * +

x = - 1.04 * 0.65 + 20.20

= 19.52

x = 19.52

b ) P(-z <  Z < z) = 70%

P(-z <  Z < z) = 0.70

P(Z < z) - P(Z < z) = 0.70
2P(Z < z) - 1 = 0.70
2P(Z < z) = 1 + 0.70
2P(Z < z) = 1.70
P(Z <,z) = 1.70 / 2 = 0.85
P(Z <  z) = 0.85
z = - 1.04

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 0.65 + 20.20

= 20.876

x = 20.88

Using z-score formula,

x = z * +

x = -1.04 * 0.65 + 20.20

= 19.52

x = 19.52