Question

note: for any part below, if it cannot be answered, enter dne. read the directions carefully and pay attention to the different sample sizes. in this problem, round to 4 decimal places. a quality control company was hired to study the length of meter sticks produced by a certain company. the team carefully measured the length of many many meter sticks, and the distribution seems to be slightly skewed to the right with a mean of 99.84 cm and a standard deviation of 0.18 cm. a) what is the probability of finding a meter stick with a length of more than 100.04 cm? b) what is the probability of finding a group of 32 meter sticks with a mean length of less than 99.81 cm? c) what is the probability of finding a group of 46 meter sticks with a mean length of more than 99.86 cm? d) what is the probability of finding a group of 28 meter sticks with a mean length of between 99.81 and 99.86 cm? e) for a random sample of 36 meter sticks, what mean length would be at the 92nd percentile?

Explanation:

Step1: Calculate z - score for part a

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(\mu = 99.84\), \(\sigma=0.18\), and \(x = 100.04\).

$$z=\frac{100.04 - 99.84}{0.18}=\frac{0.2}{0.18}\approx1.1111$$

We want \(P(X>100.04)\), which is \(P(Z > 1.1111)\). Using the standard normal table \(P(Z>z)=1 - P(Z\leq z)\). \(P(Z\leq1.11)=0.8665\), \(P(Z\leq1.12)=0.8686\). Using linear - interpolation (or a calculator), \(P(Z\leq1.1111)\approx0.8670\). So \(P(Z > 1.1111)=1 - 0.8670 = 0.1330\)

Step2: Calculate z - score for part b

For the sampling distribution of the sample mean \(\bar{x}\), the formula for the z - score is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Here, \(\mu = 99.84\), \(\sigma = 0.18\), \(n = 32\), and \(\bar{x}=99.81\)

$$z=\frac{99.81-99.84}{\frac{0.18}{\sqrt{32}}}=\frac{- 0.03}{\frac{0.18}{5.6569}}\approx\frac{-0.03}{0.0318}\approx - 0.9434$$

We want \(P(\bar{X}<99.81)\), which is \(P(Z < - 0.9434)\). Using the standard normal table, \(P(Z < - 0.94)=0.1736\), \(P(Z < - 0.95)=0.1711\). Using linear - interpolation (or a calculator), \(P(Z < - 0.9434)\approx0.1723\)

Step3: Calculate z - score for part c

For \(n = 46\), \(\mu = 99.84\), \(\sigma = 0.18\), and \(\bar{x}=99.86\)

$$z=\frac{99.86 - 99.84}{\frac{0.18}{\sqrt{46}}}=\frac{0.02}{\frac{0.18}{6.7823}}\approx\frac{0.02}{0.0265}\approx0.7547$$

We want \(P(\bar{X}>99.86)\), which is \(P(Z>0.7547)\). \(P(Z\leq0.75)=0.7734\), \(P(Z\leq0.76)=0.7764\). Using linear - interpolation (or a calculator), \(P(Z\leq0.7547)\approx0.7746\). So \(P(Z > 0.7547)=1 - 0.7746=0.2254\)

Step4: Calculate z - scores for part d

For \(n = 28\), \(\mu = 99.84\), \(\sigma = 0.18\)
For \(\bar{x}_1 = 99.81\):

$$z_1=\frac{99.81 - 99.84}{\frac{0.18}{\sqrt{28}}}=\frac{-0.03}{\frac{0.18}{5.2915}}\approx\frac{-0.03}{0.0340}\approx - 0.8824$$

For \(\bar{x}_2 = 99.86\):

$$z_2=\frac{99.86 - 99.84}{\frac{0.18}{\sqrt{28}}}=\frac{0.02}{\frac{0.18}{5.2915}}\approx\frac{0.02}{0.0340}\approx0.5882$$

We want \(P(99.81<\bar{X}<99.86)=P(-0.8824 < Z < 0.5882)\)
\(P(Z < 0.5882)\approx0.7217\), \(P(Z < - 0.8824)\approx0.1894\)
\(P(-0.8824 < Z < 0.5882)=0.7217-0.1894 = 0.5323\)

Step5: Calculate the value for part e

The z - score corresponding to the 92nd percentile is \(z = 1.4051\) (using a standard normal table or calculator).
Using the formula \(\bar{x}=\mu+z\frac{\sigma}{\sqrt{n}}\), with \(\mu = 99.84\), \(\sigma = 0.18\), \(n = 36\)

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Answer:

a) \(0.1330\)
b) \(0.1723\)
c) \(0.2254\)
d) \(0.5323\)
e) \(99.8822\)