Question

According to a recent​ study, the average length of a newborn baby is 19.219.2 inches with a standard deviation of 1.21.2 inch. The distribution of lengths is approximately Normal. Complete parts​ (a) through​ (c) below. Include a Normal curve for each part.

A.The probability that a baby will have a length of 20.4 inches or more is
B. The probability that a baby will have a length of 21.5 inches or more
C. The probability that a baby will have a length between 17.6 and 20.8 inches

Answer 1

a) 0.159

b) 0.0281

c) 0.816

Explanation:

We are given the following information in the question:

Mean, μ = 19.2 inches

Standard Deviation, σ = 1.2 inch

We are given that the distribution of lengths is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

a) P(baby will have a length of 20.4 inches or more)

`P(x \geq 20.4)`

`P( x \geq 20.4) = P( z \geq \displaystyle(20.4 - 19.2)/(1.2)) = P(z \geq 1)`

`= 1 - P(z < 1)`

Calculation the value from standard normal z table, we have,

`P(x \geq 20.4) = 1 - 0.841 = 0.159 = 15.9\%`

b) P( baby will have a length of 21.5 inches or more)

`P(x \geq 21.5) = P(z \geq \displaystyle(21.5-19.2)/(1.2)) = P(z \geq 1.91)\\\\P( z \geq 1.91) = 1 - P(z < 1.91)`

Calculating the value from the standard normal table we have,

`1 - 0.972 = 0.0281 = 2.81\%\\P( x \geq 21.5) = 2.81\%`

c)P(baby will have a length between 17.6 and 20.8 inches)

`P(17.6 \leq x \leq 20.8) = P(\displaystyle(17.6 - 19.2)/(1.2) \leq z \leq \displaystyle(20.8-19.2)/(1.2)) = P(-1.33 \leq z \leq 1.33)\\\\= P(z \leq 1.33) - P(z < -1.33)\\= 0.908 - 0.092 = 0.816 = 81.6\%`

`P(17.6 \leq x \leq 20.8) = 81.6\%`