A tire manufacturer wants to estimate the average number of miles that may be driven in a tire of a certain type before the tire wears out. Assume the population is normally distributed. A random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded, find the 97% confidence interval using the sample data.

97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].

Step-by-step explanation:

We are given that a random sample of tires is chosen and are driven until they wear out and the number of thousands of miles is recorded;

32, 33, 28, 37, 29, 30, 22, 35, 23, 28, 30, 36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                                P.Q.  =    ~

where, = sample average number of miles = = 30.25

            s  = sample standard deviation = = 4.71

            n = sample of tires = 12

             = population average number of miles

Here for constructing a 97% confidence interval we have used One-sample t-test statistics as we don't know about population standard deviation.

So, 97% confidence interval for the population mean, is ;

P(-2.55 < < 2.55) = 0.97  {As the critical value of t at 11 degrees of

                                              freedom are -2.55 & 2.55 with P = 1.5%}  

P(-2.55 < < 2.55) = 0.97

P( < < ) = 0.97

P( < < ) = 0.97

97% confidence interval for = [ , ]

                                        = [ , ]

                                        = [26.78 miles, 33.72 miles]

Therefore, 97% confidence interval for the average number of miles that may be driven is [26.78 miles, 33.72 miles].

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1