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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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Answer:

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Step-by-step explanation:

Given:

An exponential function is of the form y=ab^{x}, where, a\ne 0.

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Let us check each function now.

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Option 6: A=1700(0.93)^{t}

Here, a = 1700, b = 0.93

As 0.93 < 1, the function is exponential decay.

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