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Aleksandr [31]
2 years ago
8

In 1980 (before the era of in-vitro fertilization), there were 3,612,258 births in the United States (according to the U.S. Cens

us Bureau), of which 68,339 births were to twins and 1,337 births were to triplets or more.1. What was the probability of having multiples (twins or more) in 1980?
Mathematics
1 answer:
scoray [572]2 years ago
8 0

Answer:

There was a 1.93% probability of having multiples (twins or more) in 1980.

Step-by-step explanation:

There were 3,612,258 births in the US in 1980.

Of those,

68339 + 1337 = 69676 were multiplies.

So

What was the probability of having multiples (twins or more) in 1980?

P = \frac{69676}{3612258} = 0.0193

There was a 1.93% probability of having multiples (twins or more) in 1980.

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The waking time of a student under common conditions is normally distributed with mean of 30 hours and a standard deviation of 5
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Answer:

If the walking time is greater than or equal to 38.225 hours, than it exceeds 95% probability that is lie in top 5%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 30 hours

Standard Deviation, σ = 5 hours

We are given that the distribution of waking time is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.95

P( X < x) = P( z < \displaystyle\frac{x - 30}{5})=0.95  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 30}{5} = 1.645\\\\x = 38.225  

Thus, if the walking time is greater than or equal to 38.225 hours, than it exceeds 95% probability that is lie in top 5%.

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