For budgeting purposes, the vehicle maintenance manager for a municipal water and sewer utility needs to know how long the brakes on service trucks will last (in miles) before requiring replacement. the maintenance records for 20 trucks indicate that the mean life is 28,536 miles with a standard deviation of 861 miles. based on this sample, construct a 95% confidence interval for the true population mean for brake life.

[ 28,133, 28,939 ]

Explanation:

Given:

Number of trucks, n = 20

Mean life, μ = 28,536 miles

Standard deviation = 861 miles

Confidence level of interval = 95%

Now,

degree of freedom, df = n - 1

or

df = 20 - 1

or

df = 19

From the TINV function

t* = TINV ( P , df )

here P = 0.05 for 95% confidence

t* = TINV ( 0.05 , 19 )

or

t* = 2.093

Therefore,

The Margin of error =

on substituting the respective values, we have

The Margin of error =  

or

The margin of error = 402.955 ≈ 403

Hence,

For 95% confidence, The interval for population mean  will be

= Mean ± Margin of error

= 28,536 ± 403

or

the interval will be

= [ 28,536 - 403 , 28,536 + 403 ]

= [ 28,133, 28,939 ]

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂