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balandron [24]
3 years ago
5

A certain paper suggested that a normal distribution with mean 3,500 grams and a standard deviation of 560 grams is a reasonable

model for birth weights of babies born in Canada.
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?
Mathematics
1 answer:
Natalka [10]3 years ago
5 0

Answer: the probability that a randomly selected Canadian baby is a large baby is 0.19

Step-by-step explanation:

Since the birth weights of babies born in Canada is assumed to be normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = birth weights of babies

µ = mean weight

σ = standard deviation

From the information given,

µ = 3500 grams

σ = 560 grams

We want to find the probability or that a randomly selected Canadian baby is a large baby(weighs more than 4000 grams). It is expressed as

P(x > 4000) = 1 - P(x ≤ 4000)

For x = 4000,

z = (4000 - 3500)/560 = 0.89

Looking at the normal distribution table, the probability corresponding to the z score is 0.81

P(x > 4000) = 1 - 0.81 = 0.19

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An IQ test is designed so that the mean is 100 and the standard deviation is 18 for the population of normal adults. Find the sa
melisa1 [442]

Answer:

At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

In this question:

We need a sample size of at least n.

n is found when M = 3. We have that \sigma = 18.

M = z*\frac{\sigma}{\sqrt{n}}

3 = 1.645*\frac{18}{\sqrt{n}}

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Rounding up,

At least 98 people need to be sampled. It is not a huge number, that is, it is not difficult to sample 98 people, so it is a reasonable sample size for a real world calculation.

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