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Rina8888 [55]
3 years ago
9

Find the probability that an SRS of 36 babies will have an average birthweight of over 3.9 kg. Write your answer as a decimal. R

ound your answer to two places after the decimal.
Mathematics
1 answer:
Jlenok [28]3 years ago
3 0

Complete question :

The birthweight of newborn babies is Normally distributed with a mean of 3.96 kg and a standard deviation of 0.53 kg. Find the probability that an SRS of 36 babies will have an average birthweight of over 3.9 kg. Write your answer as a decimal. Round your answer to two places after the decimal

Answer:

0.75151

Step-by-step explanation:

Given that :

Mean weight (m) = 3.96kg

Standard deviation (σ) = 0.53kg

Sample size (n) = 36

Probability of average weight over 3.9

P(x > 3.9)

Using the z relation :

Zscore = (x - m) / (σ / √n)

Zscore = (3.9 - 3.96) / (0.53 / √36)

Zscore = - 0.06 / 0.0883333

Zscore = −0.679245

Using the Z probability calculator :

P(Z > - 0.679245) = 0.75151

= 0.75151

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(4x10^2) - (2x10^-1)
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Answer:

Step-by-step explanation:

The LCD here is 10.

Multiplying numerator and denominator of (4*10^2) by 10 results in:

40*10^3

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     10

and rewriting (2x10^-1) with positive exponent in the denominator results in:

   2

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So, before simplification, this difference comes out to be:

40000 - 2        39998            19999

----------------- = ---------------- or ------------

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3 years ago
Propane has a heat energy of 2,470 BTU/ft cubed. How much energy can a propane tank with a volume of 43.7 ft cubed produce?
professor190 [17]

Answer: 107939\ BTU

Step-by-step explanation:

Given

Propane produced heat energy of 2470\ BTU/ft^3

The volume of the tank is 43.7\ ft^3

The energy produced by this propane tank is

\Rightarrow 2470\times 43.7=107939\ BTU

4 0
2 years ago
Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of
IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable <em>X</em> following a Binomial distribution with parameters <em>n </em>and <em>p</em>.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  • np ≥ 10
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The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.10=1

Thus, a Normal approximation to binomial can be applied for population 1, if <em>n</em> = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided <em>n</em> as follows:

n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10

Thus, a Normal approximation to binomial can be applied for population 2, if <em>n</em> = 100, 50, 40 and 20.

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Answer:

Step-by-step explanation:

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8x - 54 = 82 simplify​
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Answer:

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Step-by-step explanation:

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