Sebastian opened a savings account with $800 that pays no interest. He deposits an

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

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

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

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

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So

X = 35.5

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

$Z = \frac{35.5 - 26.751}{5.08}$

$Z = 1.72$

$Z = 1.72$ has a pvalue of 0.9573

X = 24.5

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

$Z = \frac{24.5 - 26.751}{5.08}$

$Z = -0.44$

$Z = -0.44$ has a pvalue of 0.3300

0.9573 - 0.3300 = 0.6273

0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday.

(d) more than 40 will live beyond their 90th birthday

This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

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

$Z = \frac{40.5 - 26.751}{5.08}$

$Z = 2.71$

$Z = 2.71$ has a pvalue of 0.9966

1 - 0.9966 = 0.0034

0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

6 0

3 years ago

A triangle has side lengths of 10 cm, 24 cm, and 32 cm. Classify the triangle as acute, obtuse, right of not a triangle.

Stolb23 [73]

It would be obtuse.

8 0

4 years ago

Read 2 more answers

How many significant figures are in the number 0.0905?

algol [13]

Answer:

Number of Significant Figures: 3

The Significant Figures are 9 0 5

Step-by-step explanation:

7 0

3 years ago

The distance of an object thrown upward is modeled by the equation h=-2x^2+3t+4, where h is the height of the object in feet at

kipiarov [429]

So for this, we will have to find the zeros (x-intercepts) of the equation. In this case, we will be using the quadratic formula, which is $x=\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a}$ (a = x^2 coefficient, b = x coefficient, c = constant). Using our info, our equation is such: $x=\frac{-3+\sqrt{3^2-4*(-16)*4}}{2*(-16)},\frac{-3-\sqrt{3^2-4*(-16)*4}}{2*(-16)}$

Firstly, solve the multiplications and the exponents: $x=\frac{-3+\sqrt{9+256}}{-32},\frac{-3-\sqrt{9+256}}{-32}$

Next, do the addition: $x=\frac{-3+\sqrt{265}}{-32},\frac{-3-\sqrt{265}}{-32}$

Next, plug in the equations into the calculator and your answer will be: $x=-0.41,0.60$

Since we can't have negative time in this situation, it can't be -0.41 seconds. Which means that the object is visible for about 0.60 seconds, or C.

8 0

4 years ago

Ac and bd are perpendicular bisectors of each other. adc. Find eab

olasank [31]

Let ∠ ADC = 2β

Ac and BD are perpendicular bisectors of each other ⇒⇒ (Given information)

∴ BD bisects the angle ADC
∴ ∠ADE = 0.5 ∠ADC = β

And in ΔADE:
∵∠DEA = 90° ⇒⇒⇒ from the given information
∴∠DAE = 90° - β

And AC bisects ∠DAB ⇒⇒⇒ from the given information
∴∠EAB = ∠DAE = 90° - β

6 0

4 years ago

Read 2 more answers