Question

TotCo is developing a new deluxe baby bassinet. If the length of a newborn baby is normally distributed with a mean of 50 cm and a standard deviation of 5 cm, what should be the interior length of the bassinet to ensure that 99 percent of newborn babies will fit, with a safety margin of 15 cm on each end of the bassinet?

Answer 1

91.63 cm is the interior length of the bassinet to ensure that 99 percent of newborn babies will fit, with a safety margin of 15 cm on each end of the bassinet.

Explanation:

We are given the following information in the question:

Mean, μ = 50 cm

Standard Deviation, σ = 5 cm

We are given that the distribution of length of a newborn baby is a bell shaped distribution that is a normal distribution.

Formula:

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

P(X<x) = 0.99

We have to find the value of x such that the probability is 0.99

P(X < x)

`P( X < x) = P( z < \displaystyle(x - 50)/(5))=0.99`

Calculation the value from standard normal table, we have,

`P(z<2.326) = 0.99`

`\displaystyle(x - 50)/(5) = 2.326\\x = 61.63`

Thus, 99% of newborn babies will have a length of 61.63 cm or less.

There is a safety margin of 15 cm on each end of the bassinet

Length of bassinet =

`61+63 + 15 +15 = 91.63\text{ cm}`