Question

for questions 7 - 9, refer to the situation
the lengths of babies born at ust hospital in the last year were normally distributed. the mean length was 55 cm and the standard deviation was 2.25 cm.

1. what percent of the babies born at the hospital are between 52.75 cm and 59.5 cm?

a. 81.5%
b. 83.85%
c. 95%
d. 99.7%

1. what percent of the babies are shorter than 57.25 cm?

a. 84%
b. 81.5%
c. 99.85%
d. 95%

1. if there were 12000 babies born in the said hospital, how many babies are between 52.75 cm and 57.25 cm?

a. 4080
b. 8160
c. 9780
d. 5700

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 55\) (mean) and \(\sigma=2.25\) (standard deviation).
For \(x = 52.75\):
\(z_1=\frac{52.75 - 55}{2.25}=\frac{- 2.25}{2.25}=-1\)
For \(x = 59.5\):
\(z_2=\frac{59.5 - 55}{2.25}=\frac{4.5}{2.25}=2\)
Using the standard normal distribution table, \(P(Z < - 1)=0.1587\) and \(P(Z < 2)=0.9772\)
\(P(-1<Z<2)=P(Z < 2)-P(Z < - 1)=0.9772-0.1587 = 0.8185\approx81.5\%\)

Step2: For question 8

For \(x = 57.25\)
\(z=\frac{57.25 - 55}{2.25}=\frac{2.25}{2.25}=1\)
Using the standard normal distribution table, \(P(Z < 1)=0.8413\approx84\%\)

Step3: For question 9

For \(x = 52.75,z=-1\) and for \(x = 57.25,z = 1\)
\(P(-1<Z<1)\)
Using the standard normal distribution table, \(P(Z < 1)=0.8413\) and \(P(Z < - 1)=0.1587\)
\(P(-1<Z<1)=0.8413-0.1587 = 0.6826\)
If \(N = 12000\), then the number of babies \(n=12000\times0.6826 = 8191.2\approx8160\) (due to rounding in z - table values)

Answer:

1. A. \(81.5\%\)
2. A. \(84\%\)
3. B. \(8160\)