Question

a large manufacturer makes large batches of chocolate chip cookes to be packaged. 95% of the cookies have between 16 and 20 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 12 (1 point) a normal distribution has a mean of 98 and a standard deviation of 6. find the z - score for a data value of 96. round your answer to two decimal places. your answer: answer question 13 (bonus) (1 point) match each z - score to the corresponding percentile given from the standard normal table.

Explanation:

Step1: Use the property of normal distribution

In a normal distribution, about 95% of the data lies within \( \mu\pm2\sigma \). Let \( \mu \) be the mean and \( \sigma \) be the standard deviation. We have the equations \( \mu - 2\sigma=16 \) and \( \mu + 2\sigma=20 \).

Step2: Solve the system of equations for the mean

Add the two equations \( (\mu - 2\sigma)+(\mu + 2\sigma)=16 + 20 \).

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Step3: Solve the system of equations for the standard deviation

Substitute \( \mu = 18 \) into \( \mu+2\sigma = 20 \). Then \( 18+2\sigma=20 \).

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Step4: Calculate the z - score

The formula for the z - score is \( z=\frac{x-\mu}{\sigma} \). Given \( x = 96\), \( \mu=98\), and \( \sigma = 6\).

$$ z=\frac{96 - 98}{6}=\frac{-2}{6}\approx - 0.33 $$

Answer:

Mean: \(18\) chocolate chips
Standard Deviation: \(1\) chocolate chip
For the z - score: \(-0.33\)