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3 years ago

Please answer number 5 find the surface area

Mathematics

1 answer:

Margaret [11]3 years ago

3 0

Ill edit this later to answer the question, but what is the question???

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3 years ago

On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas,

erastova [34]

Answer:

The distribution of X is normal

$P(X < 1.4) = 0.12654$

$P(X > 3.5) = 0.36051$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 3 \ hours$

The standard deviation is $\sigma = 1.4 \ hours$

Generally given from the question that the amount of time spent alone by the population size is normally distributed then then the distribution of X (i.e the amount of time spent by the sample size (the one Mercurian)) will be normally distributed

Generally the probability that the child spend less than one hour in a day is mathematically represented as

$P(X < 1.4) = P(\frac{X - \mu}{\sigma} < \frac{1.4 - \mu}{\sigma} )$

Here $\frac{X - \mu}{\sigma } = Z (The\ standardized\ value\ of\ X)$

$P(X < 1.4) = P(Z < \frac{1.4 - 3.0}{1.4} )$

$P(X < 1.4) = P(Z < -1.1429 )$

From the z-table the value of

$P(Z < -1.1429 )=0.12654$

So $P(X < 1.4) = 0.12654$

Generally the percentage of children that spends over 3.5 hours unsupervised is mathematically represented as

$P(X > 3.5) = P(\frac{X - \mu}{\sigma} > \frac{3.5 - \mu}{\sigma} )$

$P(X > 3.5) = P(Z > \frac{3.5 - 3.0}{1.4} )$

$P(X > 3.5) = P(Z > 0.3571 )$

From the z-table the value of

$P(Z >0.3571 )=0.36051$

So $P(X > 3.5) = 0.36051$

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3 years ago