Question

a large manufacturer makes large batches of chocolate chip cookes to be packaged. 68% of the cookies have between 13 and 17 chocolate chips per cookie. assuming this data for the number of chocolate chips in a cookie follows a normal distribution find the mean and standard deviation of that distribution. give exact results. mean: chocolate chips standard deviation: chocolate chips. question 8 (1 point) a machine fills 10lb bags of sugar. 99.7% of the bags have weights between 9.895 lbs and 10.405 lbs. assuming this data for the weight of the bags follows a normal distribution find the mean and standard deviation of that distribution. give exact results. mean: lbs standard deviation: lbs. question 9 (1 point) give an exact answer based on the empirical rule for the question given below. the price paid for a particular model of hd television is approximately a normal distribution. the mean price paid is $1400 and the standard deviation is $135. what is the approximate percentage of buyers who paid between $1400 and $1,535?

Explanation:

Step1: Recall the empirical rule for normal distribution

The empirical rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation (\(\sigma\)) of the mean (\(\mu\)), i.e., \(\mu - \sigma\) to \(\mu+\sigma\)
• Approximately \(95\%\) of the data lies within \(2\) standard deviations of the mean, i.e., \(\mu - 2\sigma\) to \(\mu + 2\sigma\)
• Approximately \(99.7\%\) of the data lies within \(3\) standard deviations of the mean, i.e., \(\mu-3\sigma\) to \(\mu + 3\sigma\)

Step2: Solve for the first problem (chocolate - chip cookies)

Given that \(68\%\) of the data lies between \(13\) and \(17\). Using the formula \(\mu-\sigma=x_1\) and \(\mu+\sigma=x_2\) (where \(x_1 = 13\) and \(x_2=17\))
We can solve the system of equations:
\(

$$\begin{cases}\mu-\sigma=13\\\mu+\sigma = 17\end{cases}$$

\)
Add the two equations: \((\mu-\sigma)+(\mu+\sigma)=13 + 17\)
\(2\mu=30\), so \(\mu=\frac{30}{2}=15\)
Substitute \(\mu = 15\) into \(\mu+\sigma=17\), we get \(15+\sigma=17\), then \(\sigma=2\)

Step3: Solve for the second problem (sugar - bags)

Given that \(99.7\%\) of the data lies between \(9.895\) and \(10.405\). Using the formula \(\mu - 3\sigma=x_1\) and \(\mu+3\sigma=x_2\) (where \(x_1 = 9.895\) and \(x_2 = 10.405\))
Add the two equations \((\mu - 3\sigma)+(\mu+3\sigma)=9.895 + 10.405\)
\(2\mu=20.3\), so \(\mu=\frac{20.3}{2}=10.15\)
Substitute \(\mu = 10.15\) into \(\mu+3\sigma=10.405\)
\(10.15+3\sigma=10.405\)
\(3\sigma=10.405 - 10.15=0.255\)
\(\sigma=\frac{0.255}{3}=0.085\)

Step4: Solve for the third problem (HD - television)

Given \(\mu = 1400\) and \(\sigma=135\), \(x_1 = 1400\) and \(x_2=1535\)
Since \(\mu+\sigma=1400 + 135=1535\)
The empirical rule states that approximately \(68\%\) of the data lies within \(\mu-\sigma\) and \(\mu+\sigma\). The data from \(\mu\) to \(\mu+\sigma\) is half of the \(68\%\) data within \(\mu-\sigma\) and \(\mu+\sigma\)
So the percentage of data from \(\mu\) to \(\mu+\sigma\) is \(\frac{68\%}{2}=34\%\)

Answer:

• For the chocolate - chip cookies: Mean \(=15\) chocolate chips, Standard Deviation \(=2\) chocolate chips
• For the sugar - bags: Mean \(=10.15\) lbs, Standard Deviation \(=0.085\) lbs
• For the HD - television: \(34\%\)