6.52

x inches

inches

==

10 500

0 005

.

.σ

a) z = − = =10 520 10 5000 005

0 0200 005

4. .

...

3.6sin%10015.05.0)520.10( TablegUxP ⇒=+=≤

b) z1

10 485 10 5000 005

3= − = −. ..

z2

10 515 10 5000 005

3= − =. ..

From Table 6.3, A = 0.4987 P x( . . ) . . .10 485 10 515 2 0 4987 0 9974 99 87%< < = × = ⇒ From Table 6.3 c) From Table 6.3,

%76.989876.04938.02)5.25.2(,5.2 ⇒=×=+≤≤−= σσ xxxPz

012.09876.01)( =−=rejectionP 6.54 a) σ = 5000 From table 6.3 for area = 0.40 z ≈ 1.28

⇒ x − = −50 000

5000128

,.

x = 43,600 miles

b) z1

60 000 50 0005000

2= − =, , area = 0.4772

z2

70 000 50 0005000

4= − =, , area = 0.5

⇒ 0.5-0.4772 = 0.0228 (100,000 tires)(0.0228) = 2280 tires fail between 60,000 to 70,000 miles

c) 20 000 50 000

50006

, ,− = −

⇒ No tires expected to have life less than 20,000 d) Major assumption: Life span of tire follow normal distribution.

10.500 10.520

50,000

10%

40%

50,00060,000 70,000

6.57a Confidence level: 95% 1-α = 0.95 ⇒α = 0.05 0.5 - α/2 = 0.475 ⇒ z = 1.96 (Table 6.3)

µ σ

µ

µ

α= ±

= ±

= ±

x zn

mph with confidence of

2

30196 2

4030 0 620 95%

( . )( )

. 6.57b For a large sample, n>30, nzx /2/ σµ α±= For 90, 95, and 99% confidence levels, xα/2 = 1.64, 1.96, and 2.58 respectively. For 90% confidence level, the confidence interval is ±1.64x0.2/(40)1/2 = ±0.052 oz. For 95% and 99% confidence levels, the intervals are ±0.061 oz and ±0.082 oz. 6.59 (a) The mean is 16.042 oz. and the sample standard deviation is 0.079 oz. The standard deviation of the mean is 0.07941/(12)1/2 = 0.0229 oz. (b) This is a t-distribution problem with ν= 12-1 = 11 and α/2 = 0.025. From the t-distribution table, t has a value of 2.201. The confidence interval on the mean is then ±tα/2S/(n)1/2 = 2.201x0.0229 = 0.0504 oz. 6.60 Find 95% confidence interval on the mean

x miles

S miles

n

===

50 000

5000

100

,

1-α = 0.95 ⇒α = 0.05 0.5-α/2 = 0.475 z = 1.96 (From Table 6.3)

µ σα= ±x z

n2

µµ

= ±= ±

50 000 196 5000 100

50 000 980

, . ( / )

, miles