Oven thermostats were tested by setting them to 350°F and

Solution 8E

Step1 of 6:

Let us consider a random variable X it represents the oven temperature. Here random variable X follows normal distribution with mean standard deviation and n = 350.

That is

The probability density function of normal distribution is given by

, .

Where,

x = random variable

= mean of X

= variance of X

= standard deviation os X

= mathematical constant and its value is 3.14

Here our goal is:

a. We need to find  90% confidence interval for the mean oven temperature.

b. We need to find 95% confidence interval for the mean oven temperature.

c. We need to find the confidence level of the interval (347.5, 348.9).

d. We need to check how many thermostats must be sampled so that a 90% confidence interval specifies the mean to within ±0.8°F.

e. We need to find how many thermostats must be sampled so that a 95% confidence interval specifies the mean to within ±0.8°F.

Step2 of 6:

a).

Here we have to find 90%  CI, let us take .

Now,

   = 0.05

Z-scores(are to the right) is given by

  is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9500 value falls, it falls in row 1.6 under column 0.04.

Hence,

90% confidence interval for the mean oven temperature is given by

(347.7529, 348.6470)

Hence,  90% confidence interval for the mean oven temperature is (347.7529, 348.6470).

Step3 of 6:

b).

Here we have to find 95%  CI, let us take .

Now,

   = 0.025

Z-scores(are to the right) is given by

  is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9750 value falls, it falls in row 1.9 under column 0.06.

Hence,

95% confidence interval for the mean oven temperature is given by

(347.6657, 348.7342)

Hence,  95% confidence interval for the mean oven temperature is (347.6657, 348.7342).

Step4 of 6:

c).

We have and n = 350.

In a given information we have upper bound 180

Now,

348.9 - 348.2 = (0.2726)

                 0.7 =  (0.2726)

                                                =

   = 2.5678

Hence, = 2.56.

Z-scores is given by , and this value is obtained from standard normal table(area under normal curve). In standard normal table we have to see in row 2.5 under column 0.06.

 = 0.9948

Area to the right of Z is  1 - 0.9948

                                      = 0.0052.

Now,

             = 0.0104

Therefore level is (1-) = 1 - 0.0104

                                       = 0.9896.

Hence level is 0.9896.

Step5 of 6:

d).

We know that , s = 5.1 .

    = 0.8

= 0.8

                              = 109.3070.

                             110

Hence, n = 110.

Step6 of 6:

e).

We know that , s = 5.1 .

    = 0.8

= 0.8

                      = 156.1250.

                 157

Hence, n = 157.