Question

according to a medical journal, the distribution of weights of newborn babies is approximately normal, with a mean of 3390 grams and a standard deviation of 550 grams. use a technology or a table to answer parts (a) through (c) below. for each include an appropriately labeled and shaded normal curve. the probability that a newborn baby will weigh more than 3900 grams is 17.6%. (round to one decimal place as needed.) b. what percentage of newborn babies weigh between 3100 and 3900 grams? select the correct graph below. oa. ob. density density 3100 3100 3900 3900 1740 3390 5040 1740 3390 5040 oc. od. density density 3100 3100 3900 3900 1740 3390 5040 1740 3390 5040

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 3390\) (mean) and \(\sigma=550\) (standard deviation).
For \(x = 3100\):
\(z_1=\frac{3100 - 3390}{550}=\frac{- 290}{550}\approx - 0.53\)
For \(x = 3900\):
\(z_2=\frac{3900 - 3390}{550}=\frac{510}{550}\approx0.93\)

Step2: Find the probabilities

Using a standard normal table or technology (e.g., a TI - 84 Plus: normalcdf\((-0.53,0.93)\)), we find the probability \(P(-0.53<Z<0.93)\).
\(P(Z < 0.93)\approx0.8238\) and \(P(Z<-0.53)\approx0.2981\)
\(P(-0.53 < Z < 0.93)=P(Z < 0.93)-P(Z<-0.53)\)
\(P(-0.53 < Z < 0.93)=0.8238 - 0.2981=0.5257\)

Answer:

The percentage of newborn babies that weigh between \(3100\) and \(3900\) grams is \(52.6\%\) (rounded to one decimal place).

For the graph:
The correct graph is \(C\) because the normal curve is centered at the mean \(\mu = 3390\), and we are interested in the area between \(x = 3100\) and \(x = 3900\). In graph \(C\), the shaded region is between \(3100\) and \(3900\) symmetrically (as much as possible given the non - symmetric z - scores) around the mean.