The table shows the profit from a school book fair based on the number of books sold.

Suppose a continuous probability distribution has an average of μ=35 and a standard deviation of σ=16. Draw 100 times at random

yulyashka [42]

Answer:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For sums, we have that the mean is $\mu*n$ and the standard deviation is $s = \sigma \sqrt{n}$

In this problem, we have that:

$\mu = 100*35 = 3500, \sigma = \sqrt{100}*16 = 160$

This probability is the pvalue of Z when X = 4000 subtracted by the pvalue of Z when X = 3000.

X = 4000

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

By the Central Limit Theorem

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

$Z = \frac{4000 - 3500}{160}$

$Z = 3.13$

$Z = 3.13$ has a pvalue of 0.9991

X = 3000

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

$Z = \frac{3000 - 3500}{160}$

$Z = -3.13$

$Z = -3.13$ has a pvalue of 0.0009

0.9991 - 0.0009 = 0.9982

So the correct answer is:

To use a Normal distribution to approximate the chance the sum total will be between 3000 and 4000 (inclusive), we use the area from a lower bound of 3000 to an upper bound of 4000 under a Normal curve with its center (average) at 3500 and a spread (standard deviation) of 160 . The estimated probability is 99.82%.

5 0

2 years ago

What is the range of the exponential function shown below

Vsevolod [243]

Answer : y>0

f(x) = 9*2^x

f(x) is an exponential function

$f(x) = 9*2^x$

When we plug in positive value for x , the value of y is positive

When we plug in negative value for x , the value y is also positive

So for any value of x, the y value is positive always.

Range is the set of y values for which the function is defined

y values are positive , so range is y >0

3 0

1 year ago

Read 2 more answers

Find the value of x, & y.<br> 3x-y+8x-7=x+2y+5x+3y

Advocard [28]

Please rewrite this equation(s) using parantheses than needed

thank you

5 0

1 year ago

preliminary sample of 100 labourers was selected from a population of 5000 labourers by simple random sampling. It was found tha

VladimirAG [237]

Answer:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$

n=369

Step-by-step explanation:

1) Notation and definitions

$X=40$ number of the selected labourers opt for a new incentive scheme.

$n=100$ random sample taken

$\hat p=\frac{40}{100}=0.4$ estimated proportion of the selected labourers opt for a new incentive scheme.

$p$ true population proportion of the selected labourers opt for a new incentive scheme.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

2) Solution tot he problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: