1. A glass tube maker claims that their tubes have lengths that are have true me

Question

1. A glass tube maker claims that their tubes have lengths that are have true mean 9.00cm, and true variance 0.25cm2 .

(a) What is the probability that a random sample of 40 tubes will have a mean less than 8.8cm?

(b) What is the probability that a random sample of 35 tubes will have a mean of more than 9.2 cm?

(c) What is the 10tth and 90th percentiles for the mean length of tubes, based on a sample of size 75?

(d) What is the probability a random sample of 75 will have a sample mean within 0.10cm of the true (pop- ulation) mean?

Explanation / Answer

A)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    8.8      
u = mean =    9      
n = sample size =    40      
s = standard deviation =    0.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.529822128      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.529822128   ) =    0.005706018 [ANSWER]

***************

b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    9.2      
u = mean =    9      
n = sample size =    35      
s = standard deviation =    0.5      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.366431913      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.366431913   ) =    0.008980239 [ANSWER]

***************

c)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    9      
z = the critical z score =    -1.281551566      
s = standard deviation =    0.5      
n = sample size =    75      
Then          
          
x = critical value = P10 =   8.926009586   [10th percentile, ANSWER]

******

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    9      
z = the critical z score =    1.281551566      
s = standard deviation =    0.5      
n = sample size =    75      
Then          
          
x = critical value =    9.073990414   [ANSWER, 90TH PERCENTILE]

**********************************

d)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound = 9-0.1 =   8.9      
x2 = upper bound = 9+0.1 =   9.1      
u = mean =    9      
n = sample size =    75      
s = standard deviation =    0.5      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -1.732050808      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    1.732050808      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.041632258      
P(z < z2) =    0.958367742      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.916735483   [ANSWER]  
  
  

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