Question

Body size in female norther fur seals, measured as total length, is approximately normally distributed with a mean of 124.6 cm and a standard deviation equal to 6.5 cm. a. about what fraction of individuals have a total body length less than 110cm? b. what fraction of female fur seals have a body length between 130 and 140cm? c. what fraction have a body length between 120 and 125 cm?

Answer 1

a)

Here, μ = 124.6, σ = 6.5 and x = 110. We need to compute P(X <= 110). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (110 - 124.6)/6.5 = -2.25

Therefore,
P(X <= 110) = P(z <= (110 - 124.6)/6.5)
= P(z <= -2.25)
= 0.0122

b)

Here, μ = 124.6, σ = 6.5, x1 = 130 and x2 = 140. We need to compute P(130<= X <= 140). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (130 - 124.6)/6.5 = 0.83
z2 = (140 - 124.6)/6.5 = 2.37

Therefore, we get
P(130 <= X <= 140) = P((140 - 124.6)/6.5) <= z <= (140 - 124.6)/6.5)
= P(0.83 <= z <= 2.37) = P(z <= 2.37) - P(z <= 0.83)
= 0.9911 - 0.7967
= 0.1944

c)

Here, μ = 124.6, σ = 6.5, x1 = 120 and x2 = 125. We need to compute P(120<= X <= 125). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (120 - 124.6)/6.5 = -0.71
z2 = (125 - 124.6)/6.5 = 0.06

Therefore, we get
P(120 <= X <= 125) = P((125 - 124.6)/6.5) <= z <= (125 - 124.6)/6.5)
= P(-0.71 <= z <= 0.06) = P(z <= 0.06) - P(z <= -0.71)
= 0.5239 - 0.2389
= 0.2850