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The quality control department at Alexander Fasteners measured the length of 100 bolts randomly selected from a specified order. The mean length was found to be 6.8 cm, and the standard deviation was 0.7 cm. The bolt lengths are normally distributed.a. Determine the percentage of bolts shorter than 5.79 cm.b. Determine the percentage of bolts longer than 7.46 cm.c. Determine the percentage of bolts shorter than 5.79 cm.d. Determine the percentage of bolts that are longer than the nominal length of 6.65 cm.

Respuesta :

Answer:

a) P [Z < 5.79]  =  7.49 %

b) P [ Z > 7.46 ] = 82.38 %

c) P [ Z > 6.65 ]  = 91.68 %

Step-by-step explanation:

Normal Distribution

Population Mean     μ₀  = 6.8 cm

Standard Deviation  of population    σ = 0,7 cm

a) P [ Z < 5.79 ]  =  ??

z value ?

z = ( 5,79 - 6,8 ) / 0,7    ⇒   z = - 1.01 / 0,7  ⇒   z = - 1.442

From z table we get:

z =  - 1.442       ⇒   P [Z < 5.79]  =  0.0749     or

P [Z < 5.79]  =  7.49 %

b)  P [ Z > 7.46 ]

z =  ( 7.46 - 6,8 ) / 0,7    ⇒  z = 0.66 / 0.7   ⇒ z = 0.942

From z table

P [ Z > 7.46 ] = 0.8238    or     P [ Z > 7.46 ] = 82.38 %

c) P [ Z > 6.65]  

z  = ( 6.65 - 6.8 ) / 0.7       ⇒   z  = - 0,15 / 0.7     ⇒ z = - 0.214

From table we get area between 6.65 and the mean, therefore we have to add  ( 0.5 ) half of total area

Then from z table  

z = - 0,214         ⇒   0,4168

Then P [ Z > 6.65 ]  = 0,4168 + 0.5

P [ Z > 6.65 ]  = 0.9168    or

P [ Z > 6.65 ]  = 91.68 %

P [ Z > 6,65 ] = 0,5 +

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