Questions: In an advertising campaign, a snack company claimed that every 18-ounce bag of its cookies contained at least 1000 chocolate chips. Two statisticians attempted to verify the claim. The accompanying data represent the number of chips in an 18-ounce bag of the company's cookies based on their study. Complete parts (a) through (e). Click here to view the chocolate chip data table. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Chocolate Chip Data Table 1085 1155 1218 1266 1345 1098 1164 1220 1268 1355 1106 1186 1221 1282 1366 1124 1189 1223 1295 1379 1127 1195 1241 1296 1399 1139 1198 1244 1296 1420 1140 1211 1246 1312 1443 1146 1215 1254 1328 1512 (d) Using the normal model from part (c), find the probability that an 18-ounce bag of chips selected at random contains at least 1000 chips. The probability is (Round to three decimal places as needed.) (e) Using the normal model from part (c), find the probability that an 18-ounce bag of chips selected at random contains between 1300 and 1400 chips. The probability is (Round to three decimal places as needed.)

In an advertising campaign, a snack company claimed that every 18-ounce bag of its cookies contained at least 1000 chocolate chips. Two statisticians attempted to verify the claim. The accompanying data represent the number of chips in an 18-ounce bag of the company's cookies based on their study. Complete parts (a) through (e). Click here to view the chocolate chip data table. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Chocolate Chip Data Table 1085 1155 1218 1266 1345 1098 1164 1220 1268 1355 1106 1186 1221 1282 1366 1124 1189 1223 1295 1379 1127 1195 1241 1296 1399 1139 1198 1244 1296 1420 1140 1211 1246 1312 1443 1146 1215 1254 1328 1512 (d) Using the normal model from part (c), find the probability that an 18-ounce bag of chips selected at random contains at least 1000 chips. The probability is (Round to three decimal places as needed.) (e) Using the normal model from part (c), find the probability that an 18-ounce bag of chips selected at random contains between 1300 and 1400 chips. The probability is (Round to three decimal places as needed.)
Transcript text: In an advertising campaign, a snack company claimed that every 18 -ounce bag of its cookies contained at least 1000 chocolate chips. Two statisticians attempted to verify the claim. The accompanying data represent the number of chips in an 18-ounce bag of the company's cookies based on their study. Complete parts (a) through (e). Click here to view the chocolate chip data table. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Chocolate Chip Data Table \begin{tabular}{|lllll|} \hline & & & & \\ 1085 & 1155 & 1218 & 1266 & 1345 \\ 1098 & 1164 & 1220 & 1268 & 1355 \\ 1106 & 1186 & 1221 & 1282 & 1366 \\ 1124 & 1189 & 1223 & 1295 & 1379 \\ 1127 & 1195 & 1241 & 1296 & 1399 \\ 1139 & 1198 & 1244 & 1296 & 1420 \\ 1140 & 1211 & 1246 & 1312 & 1443 \\ 1146 & 1215 & 1254 & 1328 & 1512 \\ \hline \end{tabular} (d) Using the normal model from part (c), find the probability that an 18 -ounce bag of chips selected at random contains at least 1000 chips. Print Done The probability is $\square$ (Round to three decimal places as needed.) (e) Using the normal model from part (c), find the probability that an 18-ounce bag of chips selected at random contains between 1300 and 1400 chips. The probability is $\square$ (Round to three decimal places as needed.)
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate the Mean and Standard Deviation

First, we need to calculate the mean (μ) and standard deviation (σ) of the number of chocolate chips in the 18-ounce bags.

Given data: \[ 1085, 1155, 1218, 1268, 1345, 1098, 1144, 1200, 1268, 1356, 1124, 1189, 1241, 1296, 1399, 1127, 1195, 1244, 1299, 1432, 1140, 1211, 1254, 1328, 1443 \]

Calculate the mean (μ): \[ \mu = \frac{\sum X}{N} = \frac{1085 + 1155 + 1218 + 1268 + 1345 + 1098 + 1144 + 1200 + 1268 + 1356 + 1124 + 1189 + 1241 + 1296 + 1399 + 1127 + 1195 + 1244 + 1299 + 1432 + 1140 + 1211 + 1254 + 1328 + 1443}{25} \] \[ \mu = \frac{30915}{25} = 1236.6 \]

Calculate the standard deviation (σ): \[ \sigma = \sqrt{\frac{\sum (X - \mu)^2}{N}} \] \[ \sigma = \sqrt{\frac{(1085-1236.6)^2 + (1155-1236.6)^2 + \ldots + (1443-1236.6)^2}{25}} \] \[ \sigma \approx 112.6 \]

Step 2: Find the Z-Score for 1000 Chips

To find the probability that an 18-ounce bag contains at least 1000 chips, we need to calculate the Z-score for 1000 chips.

\[ Z = \frac{X - \mu}{\sigma} \] \[ Z = \frac{1000 - 1236.6}{112.6} \] \[ Z \approx -2.10 \]

Step 3: Find the Probability Using the Z-Score

Using the Z-score table, find the probability corresponding to \( Z = -2.10 \).

The Z-score table gives the probability to the left of the Z-score: \[ P(Z < -2.10) \approx 0.0179 \]

Since we need the probability of at least 1000 chips: \[ P(X \geq 1000) = 1 - P(Z < -2.10) \] \[ P(X \geq 1000) = 1 - 0.0179 \] \[ P(X \geq 1000) \approx 0.9821 \]

Final Answer

The probability that an 18-ounce bag of chips contains at least 1000 chips is approximately 0.982.

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
When heated in the presence of H+, menthol reacts to form two alkenes. In the reaction above, which alkene is the major product? Circle it above and justify your answer.
failedAnswered
Graded Quiz: Pathology Graded Quiz: Pathology 0:01:51 elapsed When pathologists are examining cells under the microscope and there is an abnormality, what do they evaluate and document? A) Diagnosis, grade, stage B) Stage, agents, immunological reactions C) Diagnosis, grade, indirect causes D) Grade, stage, immunological reactions Question 7 (5 points)
failedAnswered
molecule or polyatomic ion polar or nonpolar? atom closest to negative side F2 polar nonpolar CS2 polar nonpolar CH3F polar nonpolar
failedAnswered
The goal of supply chain integration is to synchronize supply and demand of all of the organizations that are part of the chain. True False
failedAnswered
listribution for the given data. when award was won Frequency 25 37 11 9 3 9 5 9 2 1 Age (years) of Best Actress when award was won Cumulative Frequency Less than 30 Less than 40 Less than 50 Less than 60 Less than 70