Question

question 8 of 10
as an avid fisherman, nick was curious if the advertised strength of fishing line is accurate. to investigate, he used \6 pound\
fishing line to hang a bucket, which he then filled with weights until the line broke. then nick measured the total weight of the
bucket and its contents to determine the breaking strength of that piece of fishing line. after trying this with many different
sections of the fishing line, he estimated that the distribution of breaking strength for this type of line is approximately normal
with a mean of 6.44 pounds and a standard deviation of 0.75 pound.
(a) how often will this type of fishing line have a breaking strength less than
the advertised 6 pounds? (round to 2 decimal places.)
(b) what is the 60th percentile of breaking strength for this type of fishing line?
(round to 2 decimal places.)

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 6\), \(\mu=6.44\), and \(\sigma = 0.75\).

$$z=\frac{6 - 6.44}{0.75}=\frac{- 0.44}{0.75}\approx - 0.59$$

Step2: Find the probability for part (a)

Using the standard normal distribution table or a calculator with a normal - distribution function (e.g., \(P(X\lt x)=\text{normalcdf}(-\infty,x,\mu,\sigma)\) or for the standard normal \(P(Z\lt z)\)), for \(z=-0.59\), \(P(Z\lt - 0.59)=0.2776\)

Step3: Calculate the z - score for the 60th percentile for part (b)

We need to find the \(z\) - value such that \(P(Z\lt z)=0.6\). Using a standard normal table or inverse - normal function (e.g., \(z=\text{invNorm}(p,\mu = 0,\sigma = 1)\) where \(p = 0.6\)), \(z\approx0.25\)

Step4: Calculate the value of \(x\) for part (b)

Using the formula \(x=\mu+z\sigma\), with \(\mu = 6.44\), \(z = 0.25\), and \(\sigma=0.75\)

$$x=6.44+0.25\times0.75=6.44 + 0.1875=6.63$$

Answer:

a. \(27.76\%\)
b. \(6.63\) pounds