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3 years ago

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In summer, the price of strawberries goes up by $1.25 per pound. Sam bought two pounds of strawberries at the new price of $8.46

. Write an algebraic equation to determine the original price per pound (s). Use the equation to find the original price per pound.

1 answer:

morpeh [17]3 years ago

8 0

Answer:

8.46-(8.46/1.25)

Step-by-step explanation:

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A telephone exchange operator assumes that 8% of the phone calls are wrong numbers. If the operator is right, what is the probab

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Answer:

The probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

Step-by-step explanation:

Let <em>X</em> = number of phone calls that are wrong numbers.

The proportion of phone calls that are wrong numbers is, <em>p</em> = 0.08.

A sample of<em> </em><em>n</em> = 421 phone calls is selected to determine the proportion of wrong numbers in this sample.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of a Binomial distribution is:

$P(X=x)={421\choose x}0.08^{x}(1-0.08)^{421-x}$

Now, for the sample proportion to differ from the population proportion by 3% the value of the sample proportion should be:

$\hat p-p=0.03\\\hat p-0.08=0.03\\\hat p=0.11$ $\hat p-p=-0.03\\\hat p-0.08=-0.03\\\hat p=0.05$

So when the sample proportion is less than 5% or greater than 11% the difference between the sample proportion and population proportion will be greater than 3%.

If sample proportion is 5% then the value of <em>X</em> is,

$X=np=421\times 0.05=21.05\approx21$

Compute the value of P (X ≤ 21) as follows:

$P(X\leq 21)=\sum\limits^{21}_{x=0}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0106$

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$X=np=421\times 0.11=46.31\approx47$

Compute the value of P (X ≥ 47) as follows:

$P(X\geq 47)=\sum\limits^{471}_{x=47}{{421\choose x}0.08^{x}(1-0.08)^{421-x}}=0.0135$

Then the probability that the sample proportion differ from the population proportion by greater than 3% is:

$P(\hat p-p>0.03)=P(X\leq 21)+P(X\geq 47)=0.0106+0.0135=0.0241$

Thus, the probability that the sample proportion differ from the population proportion by greater than 3% is 0.0241.

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