The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. What is the probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long?What is the probability that a standard normal random variable will be between .3 and 3.2?

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

    Mean, = 30.05 inches        and    Standard deviation, = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

                  Z = ~ N(0,1)

(a) The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65)

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P( < ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

                                                                                                = 0.9985

P(X <= 30.25) = P( <= ) = P(Z <= 1) = 0.84134

Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .

(b) Let Y = Standard Normal Variable is given by N(0,1)

Which means mean of Y = 0 and standard deviation of Y = 1

So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)

 P(Y < 3.2) = P( < ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

                                                                                           = 0.99931

 P(Y <= 0.3) = P( <= ) = P(Z <= 0.3) = 0.61791

Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .

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