Question

For questions 8 and 9 : Weights of newborn babies in a certain state have normal distribution with mean 7.50lb and standard deviation 1.25lb. A newborn weighing less than 6lb is considered to be at-risk, that is, has a higher mortality rate. 8. A baby just born in this state is picked at random. The probability that the baby is at-risk is about (a) 0.885 (b) 0.933 (c) 0.115 (d) 0.067⋅ (e) 0.05 9. The hospital wants to take pictures of the newborn babies whose weight is in top 10%. The minimum weight (in lbs) required for a picture to be taken is about (a) 5.90 (b) 9.56 (c) 8.50 (d) 8.10 (e) 9.10

Answer 1

The probability that a newborn baby in the state is at-risk and the minimum birth weight required for a picture to be taken can be calculated using the normal distribution. The probability of a baby being at-risk is approximately 0.115 and the minimum birth weight required for a picture is about 8.10lb.

Step-by-step explanation:

To calculate the probability that a newborn baby in the state has a birth weight of less than 6lb, we can use the normal distribution.

The mean birth weight is 7.50lb and the standard deviation is 1.25lb.

We can use these values to calculate the z-score for a birth weight of 6lb.

z = (x - µ) / σ

z = (6 - 7.50) / 1.25 = -1.20

Next, we can use a normal distribution table or a calculator to find the probability corresponding to a z-score of -1.20. Using either method, we find that the probability is approximately 0.115, which is option (c).

For question 9, we are looking for the minimum birth weight required for a picture to be taken, which is in the top 10%.

To find this, we can use the normal distribution table or a calculator to find the z-score that corresponds to the 90th percentile.

This z-score will give us the corresponding birth weight.

z = 1.28

Next, we can solve for x using the z-score formula:

x = µ + z * σ

x = 7.50 + 1.28 * 1.25 = 8.10

Therefore, the minimum birth weight required for a picture to be taken is about 8.10lb, which is option (d).