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tensa zangetsu [6.8K]
3 years ago
6

(14.14) larry reads that 1 out of 6 eggs contains salmonella bacteria. so he never uses more than 5 eggs in cooking. if eggs do

or don't contain salmonella independent of each other, find the probability (±±0.01) that at least 1 of larry's 5 eggs contains salmonella. rn note: please be sure you read the question as 'at least 1' and not instead just find the prob of exactly one. probability is about
Mathematics
1 answer:
melomori [17]3 years ago
6 0

we know that

The probability that "at least one" is the probability of exactly one, exactly 2, exactly 3, 4 and 5 contain salmonella.

The easiest way to solve this is to recognise that "at least one" is ALL 100% of the possibilities EXCEPT that none have salmonella.

If the probability that any one egg has 1/6 chance of salmonella

then

the probability that any one egg will not have salmonella = 5/6.

Therefore

for all 5 to not have salmonella


= (5/6)^5 = 3125 / 7776

= 0.401877 = 0.40 to 2 decimal places


REMEMBER this is the probability that NONE have salmonella


Therefore

the probability that at least one does = 1 - 0.40

= 0.60


the answer is

0.60 or 60%

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The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5. C
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Answer:

The required 97.5% confidence interval is

\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm  0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\

Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces

.

Step-by-step explanation:

The amounts (in ounces) of juice in eight randomly selected juice bottles are:

15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5

Let us first compute the mean and standard deviation of the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

\bar{x} = 15.5

=STDEV(number1, number2,....)

The standard deviation is found to be

s = 0.31  

The confidence interval for the mean amount of juice in all such bottles is given by

$ \text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $\\\\

Where \bar{x} is the sample mean, n is the samplesize, s is the sample standard deviation and t_{\alpha/2} is the t-score corresponding to a 97.5% confidence level.

The t-score corresponding to a 97.5% confidence level is

Significance level = α = 1 - 0.975 = 0.025/2 = 0.0125

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.0125 and DoF = 7

t-score = 2.8412

So the required 97.5% confidence interval is

\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm  0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\

Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces.

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