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Romashka-Z-Leto [24]
1 year ago
6

A child who is trick-or-treating chooses a lollipop at random from a bowl of lollipops with mean weight 1.5 oz and standard devi

ation 0.3 oz. The child then chooses two chocolate bars at random from a bowl of bars with mean weight 2.0 oz and standard deviation 0.4 oz. The three choices are independent. Find the expected combined weight of the child's lollipop and two chocolate bars.
Mathematics
1 answer:
son4ous [18]1 year ago
3 0

Answer:

The expected combined weight of the child's lollipop and two chocolate bars is 5.5 oz.

Step-by-step explanation:

Normal probability distribution

When the distribution is normal, we use 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.

Addition of normal variables:

The mean is the sum of the means.

Find the expected combined weight of the child's lollipop and two chocolate bars.

Child lollipop: Mean 1.5 oz

Chocolate bars: Mean 2 oz

One lollipop and 2 bars combined

The mean will be 1.5 + 2*2 = 5.5 oz

The expected combined weight of the child's lollipop and two chocolate bars is 5.5 oz.

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Step-by-step explanation:

one of the catapults could have more force than the other

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How much money do winners go home with from the television quiz show Jeopardy? To determine an answer, a random sample of winner
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Answer:

The 95% confidence interval would be given by (14444.04;33657.30)

Step-by-step explanation:

Previous concepts

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".

Data: $26,650 $6,060 $52,820 $8,490 $13,660$25,840 $49,840 $23,790 $51,480 $18,960$990 $11,450 $41,810 $21,060 $7,860

We can calculate the mean and the deviation from these data with the following formulas:

\bar X= \frac{\sum_{i=1}^n x_i}{n}

s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)^2}{n-1}}

\bar X=24050.67 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=17386.13 represent the sample standard deviation

n=15 represent the sample size  

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=15-1=14

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,14)".And we see that t_{\alpha/2}=2.14

Now we have everything in order to replace into formula (1):

24050.67-2.14\frac{17386.13}{\sqrt{15}}=14444.04    

24050.67+2.14\frac{17386.13}{\sqrt{15}}=33657.30

So on this case the 95% confidence interval would be given by (14444.04;33657.30)    

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That is, the 2nd row, 3rd column entry of A is the product of the 2nd row of M_1 and the 3rd column of M_2.

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