High School

The bonding strength \( S \) of a drop of plastic glue from a particular manufacturer is thought to be well approximated by a normal distribution with mean 98 lbs and standard deviation 7.5 lbs. \( S \sim N(98, 7.5^2) \). Compute the following values using a normal model assumption.

a. What proportion of drops of plastic glue will have a bonding strength between 95 and 104 lbs according to this model?

b. A single drop of that glue had a bonding strength that is 0.5 standard deviations above the mean. What proportion of glue drops have a bonding strength that is higher?

c. What bonding strength did a drop of glue have that is at the 90th percentile?

d. What is the IQR of bonding strength for drops of glue from this manufacturer?

e. Drops of a similar plastic glue from another manufacturer (manufacturer B) is claimed to have bonding strength well approximated by a normal distribution with mean 43 kg and standard deviation of 3.5 kg \( WB.kg \sim N(43, 3.5^2) \). What is the probability that a drop of manufacturer B’s glue will have strength above the 90th percentile strength of manufacturer A’s glue? You can use the conversion: \( 1 \text{ kg} \approx 2.20462 \text{ lbs} \).

Answer :

Final answer:

To assess the distribution of glue drops' bonding strength, z-scores are used to calculate proportions within certain ranges and to determine specific percentiles, such as the 90th percentile and the interquartile range (IQR). For comparing different manufacturers' glues, the strengths are converted between pounds and kilograms before using z-scores to estimate probabilities.

Explanation:

When approximating the proportion of glue drops with a certain bonding strength using a normal distribution, we perform the following calculations:


  • To find the proportion of drops with a strength between 95 and 104 lbs, we calculate the z-scores for 95 and 104, and then use the standard normal table or a calculator to find the area between these z-scores.

  • The proportion of glue drops with a bonding strength higher than 0.5 standard deviations above the mean is found by calculating the z-score and then finding the area to the right of this z-score in the standard normal distribution.

  • To find the bonding strength at the 90th percentile, we look up the corresponding z-score and then translate this back to the strength using the mean and standard deviation.

  • The Interquartile Range (IQR) is the difference between the 75th and 25th percentile bonding strengths. We find the z-scores for these percentiles and translate them into strengths.

  • For glue from manufacturer B, we first convert the 90th percentile strength from lbs to kg, then calculate the z-score for this value using manufacturer B's mean and standard deviation, and find the probability of a higher strength value.

To address the question specifically:


  3. The proportion of drops with a bonding strength between 95 and 104 lbs needs to be calculated.

  6. For a drop of glue that had a bonding strength of 0.5 standard deviations above the mean, we look for the proportion of drops with a higher strength.

  9. We need to find the value corresponding to a bonding strength at the 90th percentile.

  12. Determining the IQR of bonding strength involves finding the values at the 25th and 75th percentiles.

  15. Finally, the probability that a drop of manufacturer B’s glue will have strength above the 90th percentile strength of manufacturer A’s glue is explored.

Learn more about Normal Distribution here:

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