suppose the lifetime of a particular brand of tire for cars is approximately normally distributed with a mean of 48,000 miles and standard deviation of 3,000 miles. step 1 of 2: what is the probability that a single tire randomly selected from the population will have a lifetime between 46,800 and 49,200 miles? round your answer to 4 decimal places if necessary.

Probability is a mathematical discipline that concerns with numerical figures of how probable an occurrence is to occur or how probably a statement is to be true. A number between 0 and 1 is the probability of an occurrence, where, broadly speaking, 0 denotes the event's impossibility and 1 denotes its certainty.

X ~ N ( µ = 48000 , σ = 3000 )

As we know, P ( 46800 < X < 49200 )

Standardizing the value, we get -

Z = ( X - µ ) / σ

Z = ( 46800 - 48000 ) / 3000

Z = -0.4

Z = ( 49200 - 48000 ) / 3000

Z = 0.4

Also,

P ( -0.4 < Z < 0.4 )

P ( 46800 < X < 49200 ) = P ( Z < 0.4 ) - P ( Z < -0.4 )

P ( 46800 < X < 49200 ) = 0.6554 - 0.3446 (Probability calculated from Z table)

P ( 46800 < X < 49200 ) = 0.3108

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