A hardware store sells 45 boxes of nails for $67.50. Assuming that each box is the same price, what is the price per box?

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

If we sample from a small finite population without replacement, the binomial distribution should not be used because the events

liraira [26]

Answer:

0.0008

Step-by-step explanation:

The formula for a hypergeometric probability is

$\frac{(_AC_x)(_BC_{n-x})}{_NC_n}$

where N is the population, A is the number of objects of type A, B is the number of objects of type B, n is the sample size and x is the number of successes.

In this problem, N is 13. A is 5, since there are 5 objects drawn, and B is 8, since there are 8 remaining objects.

The sample size, n, is 5, and x is 5:

$\frac{(_5C_5)(_8C_0)}{_{13}C_5}\\\\=\frac{1(1)}{1287}=\frac{1}{1287}\approx 0.0008$

4 0

4 years ago

Complete the place-value chart for the following number. Write its number name and tell the value of the underline digit.

daser333 [38]

Six and three-hundred-twenty four thousands. The Number 2 is in the Hundreths place value.

3 0

4 years ago

Read 2 more answers

during a 20%of sale,the sale price of an mp3 alarm clock was 35.96 what was the regular price of the radio

babymother [125]

If 20% was taken off, that means that 35.96 is 80% of the original price.
35.96 = $\frac{80}{100}$
Now, we just need to put it into fractions, cross multiply, and divide to find the original price.
$\frac{35.96}{1} = \frac{80}{100} \
$
35.96×100 = 3596
1×80 = 80
3596÷80 = 44.95
Now, to make sure, we multiply 44.95 by 20%,
44.95×0.20 = 8.99
Subtract that from 44.95
44.95-8.99 = 35.96
That is the sale price of the alarm clock.
This means that yes, the regular price of the mp3 alarm clock was
$44.95

6 0

3 years ago

A correlation where the x-value increases and the y-value increases is called a __________.

madam [21]

Positive correlation.

Zero is x increases y stays the same
Negative is both decrease

5 0

3 years ago