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 \( WB_{\text{kg}} \sim N(43, 3.5^2) \). Transform the bonding strength of manufacturer B into lbs using the conversion: \( 1 \text{ kg} \approx 2.20462 \text{ lbs} \). You can use the transformation: \( WB_{\text{lbs}} = 2.20462 \times WB_{\text{kg}} \). Compare the shape, center, and spread of the two glues' bonding strength.

(a). The proportion of drops of plastic glue with bonding strength between 95 and 104 lbs is equal to the area under the normal curve between -0.4 and 0.8.

(b). The proportion associated with this z-score, which represents the area to the right of the value.

(c). The bonding strength at the 90th percentile.

(d). The IQR is equal to the difference between the two values of x.

(e). The bonding strength of the two glues from different manufacturers compare in terms of shape, center, and spread.

(a). To find the proportion of drops of plastic glue that will have a bonding strength between 95 and 104 lbs according to this model, we need to calculate the area under the normal curve between these two values.

First, we need to standardize the values using the formula.

z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.

For 95 lbs:

z = (95 - 98) / 7.5

= -0.4

For 104 lbs:

z = (104 - 98) / 7.5

= 0.8

Next, we use a standard normal distribution table or a calculator to find the proportion associated with these z-scores. The ratio of plastic glue drops with bonding strengths between 95 and 104 lbs is equivalent to the region between -0.4 and 0.8 of the normal curve.

(b). To find the proportion of glue drops that have a bonding strength higher than 0.5 standard deviations above the mean, we need to calculate the area to the right of this value on the normal curve.

Using the standardization formula, the z-score for 0.5 standard deviations above the mean is 0.5. We then find the percentage that corresponds to this z-score, or the region to the right of the value.

(c). To find the bonding strength of a drop of glue at the 90th percentile, we need to find the value corresponding to the z-score that represents the 90th percentile.

Using a standard normal distribution table or a calculator, we can find the z-score associated with the 90th percentile, which is approximately 1.28. Using the formula z = (x - μ) / σ, we can solve for x:

1.28 = (x - 98) / 7.5

Solving for x, we find the bonding strength at the 90th percentile.

(d). The interquartile range (IQR) is a measure of the spread of the data. It represents the range between the 25th and 75th percentiles. To find the IQR of the bonding strength for drops of glue from this manufacturer, we need to find the values corresponding to the 25th and 75th percentiles.

Using a standard normal distribution table or a calculator, we can find the z-scores associated with these percentiles. Then, we use the formula

z = (x - μ) / σ to solve for x at each percentile.

The IQR is equal to the difference between the two values of x.

(e). To compare the shape, center, and spread of the two glues' bonding strength, we can convert the bonding strength of the glue from manufacturer B into lbs. Using the conversion factor

1 kg ≈ 2.20462 lbs, we can convert the mean and standard deviation of the bonding strength for manufacturer B from kg to lbs.

The transformed bonding strength follows a normal distribution with the new mean and standard deviation. We can compare the shape by looking at the bell curve of each distribution, the center by comparing the means, and the spread by comparing the standard deviations.

By analysing these factors, we can determine how the bonding strength of the two glues from different manufacturers compare in terms of shape, center, and spread.

To learn more about standard deviation from the given link.

https://brainly.com/question/475676

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