Question

The average weight of newborns is 2150 grams. The standard deviation is 480 grams. Assume the variable is normally distributed. If a newborn is selected at random, find the probability that

a) The newborn will weigh less than 2846 grams.

b) The newborn will weigh between 3118 and 4038 grams.

c) What is the 60th percentile of the weights?

d) If a sample of 40 newborns were selected, find the probability that the sample mean is greater than 3330 grams.

Answer 1

We have: $a = 2150,\,\,\sigma = 480$

a) $P(x < 2846) = P(0 < x < 2846) = \Phi \left( {\frac{{\beta - a}}{\sigma }} \right) - \Phi \left( {\frac{{\alpha - a}}{\sigma }} \right) = \Phi \left( {\frac{{2846 - 2150}}{{480}}} \right) - \Phi \left( {\frac{{0 - 2150}}{{480}}} \right) = \Phi \left( {1.45} \right) + \Phi (4.47) = 0.4265 + 0.5 = 0.9265$

Answer: 0.758

b) $P(3118 < x < 4038) = \Phi \left( {\frac{{\beta - a}}{\sigma }} \right) - \Phi \left( {\frac{{\alpha - a}}{\sigma }} \right) = \Phi \left( {\frac{{4038 - 2150}}{{480}}} \right) - \Phi \left( {\frac{{3118 - 2150}}{{480}}} \right) = \Phi \left( {3.93} \right) - \Phi \left( {2.02} \right) = 0.5 - 0.4783 = 0.0217$

Answer: 0.0217

c) We have:

$P(x < X) = 0.6 \Rightarrow P( - \infty < x < X) = 0.6 \Rightarrow \Phi \left( {\frac{{X - a}}{\sigma }} \right) + \Phi \left( \infty \right) = 0.6\\ \Rightarrow \Phi \left( {\frac{{X - a}}{\sigma }} \right) + 0.5 = 0.6 \Rightarrow \Phi \left( {\frac{{X - 2150}}{{480}}} \right) = 0.1 \Rightarrow \frac{{X - 2150}}{{480}} = 0.25 \Rightarrow X = {\rm{2270}}$

Answer: 2270

d)

$P(x > 3330) = \Phi \left( \infty \right) - \Phi \left( {\frac{{\beta - a}}{{\frac{\sigma }{{\sqrt n }}}}} \right) = 0.5 - \Phi \left( {\frac{{3330 - 2150}}{{\frac{{480}}{{\sqrt {40} }}}}} \right)\\ \approx 0.5 - \Phi \left( {15.54} \right) \approx 0.5 - 0.5 = 0$

Answer: 0