A certain paper suggested that a normal distribution with mean 3,500 grams and standard deviation
530 grams is a reasonable model for birth weights of babies born in Canada. (Round your answers to four
decimal places.)
(a) One common medical definition of a large baby is any baby that weighs more than 4,000 grams at
birth. What is the probability that a randomly selected Canadian baby is a large baby?
(b) What is the probability that a randomly selected Canadian baby weighs either less than 2,000 grams
or more than 4,000 grams at birth?
(c) What birth weights describe the 20% of Canadian babies with the greatest birth weights?
Any baby with a weight of ________ grams or more is in the greatest 20% of birth weights.
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(a) To find the probability that a randomly selected Canadian baby is a large baby (weighs more than 4,000 grams), we need to calculate the area under the normal distribution curve to the right of 4,000 grams.
Using the z-score formula: z = (x - μ) / σ
where z is the z-score, x is the value we are interested in (4,000 grams in this case), μ is the mean (3,500 grams), and σ is the standard deviation (530 grams).
z = (4000 - 3500) / 530
z = 0.9434
Next, we can use a standard normal distribution table or a calculator to find the probability associated with a z-score of 0.9434.
P(Z > 0.9434) = 1 - P(Z < 0.9434)
Using a standard normal distribution table or a calculator, we find that P(Z < 0.9434) = 0.8271.
Therefore, P(Z > 0.9434) = 1 - 0.8271 = 0.1729.
The probability that a randomly selected Canadian baby is a large baby (weighs more than 4,000 grams) is 0.1729.
(b) To find the probability that a randomly selected Canadian baby weighs either less than 2,000 grams or more than 4,000 grams at birth, we need to calculate the area under the normal distribution curve to the left of 2,000 grams and the area to the right of 4,000 grams and then add these probabilities.
Using the z-score formula:
z1 = (2000 - 3500) / 530
z1 = -2.8302
z2 = (4000 - 3500) / 530
z2 = 0.9434
Using a standard normal distribution table or a calculator, we find that:
P(Z < -2.8302) = 0.0022
P(Z > 0.9434) = 0.1729
Therefore, the probability that a randomly selected Canadian baby weighs either less than 2,000 grams or more than 4,000 grams at birth is 0.0022 + (1 - 0.1729) = 0.8293.
(c) To find the birth weights that describe the 20% of Canadian babies with the greatest birth weights, we need to find the z-score associated with the 80th percentile.
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to the 80th percentile is approximately 0.8416.
Using the z-score formula:
z = (x - μ) / σ
0.8416 = (x - 3500) / 530
Solving for x:
0.8416 * 530 + 3500 = x
446.728 + 3500 = x
3946.728 = x
Therefore, any baby with a weight of 3,946.728 grams or more is in the greatest 20% of birth weights.
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