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3 years ago

For f(x) = x^2, find f(4 - sqrt(2))

1 answer:

4 0

$\bf f(x)=x^2\qquad f(4-\sqrt{2})\implies x=4-\sqrt{2}
\\\\\\
f(4-\sqrt{2})=\left( 4-\sqrt{2} \right)^2\implies f(4-\sqrt{2})=\left( 4-\sqrt{2} \right)\left( 4-\sqrt{2} \right)$

do a FOIL on it to get the product, or just use the binomial theorem to expand the binomial and simplify

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The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de

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Answer:

1. 15.87%

2. 6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

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

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

$P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)$

$= 1 - P(z \leq 1)$

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

$P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

$\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8$

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

$P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)$

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

$P(x < 6) =0.0228 = 2.28\%$

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

$P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%$

84% of newborn babies weigh between 6.7 and 9.5 pounds.

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4 years ago