# Answer to Question #2374 in Statistics and Probability for bx

Question #2374

The compressive strength of cement is assumed to be normally distributed with a mean of 8000 kilograms per sq cm and a standard deviation of 200 kilograms per sq. cm.

a) Find the probability that the strength is less than 8200 kg per sq. cm

b) Find the probability that the strength is between 7500 and 7700 kg per sq. cm

c) Determine the value for which the probability that the strength of cement is below this value is 90%.

a) Find the probability that the strength is less than 8200 kg per sq. cm

b) Find the probability that the strength is between 7500 and 7700 kg per sq. cm

c) Determine the value for which the probability that the strength of cement is below this value is 90%.

Expert's answer

Let X be the compressive strength of cement.

Then Z = (X-8000)/200 has standard normal distribution.

a) P(X<8200) =

& = P( (X-8000)/200 < (8200-8000)/200 )=

= P(Z<1)= 0.841345

& b) P(7500 < X< 7700)& = P( (7500-8000)/200 <Z< (7700-8000)/200 )=

P( -2.5<Z< -1.5 )= F(-1.5)-F(-2.5)= F(2.5)-F(1.5)=

= 0.99379 - 0,933193 = 0.060598

c) We have to find such& S& that

P(X<S) = 0.9

Thus

0.9 = P(X<S) = P(Z < (S-8000)/200 ) =

F( (S-8000)/200).

Then, using known values of F we obtain that

(S-8000)/200 = 1.281552

Hence S = 1.281552*200+8000=8256.31

Then Z = (X-8000)/200 has standard normal distribution.

a) P(X<8200) =

& = P( (X-8000)/200 < (8200-8000)/200 )=

= P(Z<1)= 0.841345

& b) P(7500 < X< 7700)& = P( (7500-8000)/200 <Z< (7700-8000)/200 )=

P( -2.5<Z< -1.5 )= F(-1.5)-F(-2.5)= F(2.5)-F(1.5)=

= 0.99379 - 0,933193 = 0.060598

c) We have to find such& S& that

P(X<S) = 0.9

Thus

0.9 = P(X<S) = P(Z < (S-8000)/200 ) =

F( (S-8000)/200).

Then, using known values of F we obtain that

(S-8000)/200 = 1.281552

Hence S = 1.281552*200+8000=8256.31

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