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1 year ago

What proportion of these candy bars have more than 300 calories per serving? (Round your answer to two decimal places.)

Mathematics

1 answer:

Hatshy [7]1 year ago

3 0

Answer: 0.12

Step-by-step explanation:

There are 65 candy bars. Out of that;

2 candy bars have 300 to 350 calories

1 candy bar has 350 to 400 calories

4 candy bars have 400 to 450 calories

1 candy bar has 450 to 500 calories

The total proportion out of 65 candy bars that have over 300 calories is;

= ( 2 + 1 + 4 + 1) / 65

= 8/65

= 0.12

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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:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

And replacing into equation (b) the values from part a we got:

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

And rounded up we have that n=369

8 0

1 year ago

An Olympic athlete of mass 80kg competes in a 100m race. What is the best estimate of his mean kinetic energy during the race?

kipiarov [429]

For this question, you need to know that the average velocity of a 100m Olympic athlete is 10 m/s before you start solving the equation. The equation is K (energy in Jules) = 1/2 mv^2 (where m is the mass in kg and v the speed in meters per second).
The correct answer is 4000J. To see how I got it, take a look at the attachment.