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Phoenix [80]
3 years ago
13

Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig

hing less than 2500 grams at birth are considered "low weight." If we randomly select a baby, what is the probability that it has a low birth weight?
Mathematics
1 answer:
Anon25 [30]3 years ago
3 0

Answer:

3.84% probability that it has a low birth weight

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3466, \sigma = 546

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{2500 - 3466}{546}

Z = -1.77

Z = -1.77 has a pvalue of 0.0384

3.84% probability that it has a low birth weight

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On day t=0t=0t, equals, 0, the stock is at its average value of {\$}3.47$3.47dollar sign, 3, point, 47 per share, but 91.2591.25
Norma-Jean [14]

Answer:

S(t) = a.sin (b.t) + d

a = -1.5, b = (2π/365), d = 3.47

S(t) = -1.5 sin (2πt/365) + 3.47

Step-by-step explanation:

Complete Question is presented in the attached image to this solution.

- Dingane has been observing a certain stock for the last few years and he sees that it can be modeled as a function S(t) of time t (in days) using a sinusoidal expression of the form

S(t) = a.sin(b.t) + d.

On day t = 0, the stock is at its average value of $3.47 per share, but 91.25 days later, its value is down to its minimum of $1.97.

Find S(t). t should be in radians.

S(t) =

Solution

S(t) = a.sin(b.t) + d.

At t = 0, S(t) = $3.47

S(0) = a.sin(b×0) + d = a.sin 0 + d = 3.47

Sin 0 = 0,

S(t=0) = d = 3.47.

At t = 91.25 days, S(t) = $1.97

But, it is given that T has to be in radians, for t to be in radians, the constant b has to convert t in days to radians.

Hence, b = (2π/365)

S(91.25) = 1.97 = a.sin(b×91.25) + d

d = 3.47 from the first expression

S(t = 91.25) = a.sin (91.25b) + 3.47 = 1.97

1.97 = a.sin (2π×91.25/365) + 3.47

1.97 = a sin (0.5π) + 3.47

Sin 0.5π = 1

1.97 = a + 3.47

a = -1.5

Hence,

S(t) = a.sin (b.t) + d

a = -1.5, b = (2π/365), d = 3.47

S(t) = -1.5 sin (2πt/365) + 3.47

Hope this Helps!!!

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Step-by-step explanation:

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