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

Answer:

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.

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X<x) = 0.99

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

P(X < x)  

$P( X < x) = P( z < \displaystyle\frac{x - 50}{5})=0.99$  

Calculation the value from standard normal table, we have,  

$P(z$

$\displaystyle\frac{x - 50}{5} = 2.326\\x = 61.63$  

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 =

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