The diameters of pencils produced by a certain machine are normally distributed

The probability that the diameter of a randomly selected pencil will between 0.21 and 0.29 inches is 0.15865.

Step-by-step explanation:

We are given that the diameters of pencils produced by a certain machine are normally distributed with a mean of 0.30 and a standard deviation of 0.01.

Let X = diameters of pencils produced by a certain machine

SO, X ~ N()

The z-score probability distribution is given by ;

                 Z = ~ N(0,1)

where, = mean diameter = 0.30

            = standard deviation = 0.01

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, the probability that the diameter of a randomly selected pencil will between 0.21 and 0.29 inches is given by = P(0.21 inches < X < 0.29 inches)

  P(0.21 < X < 0.29) = P(X < 0.29) - P(X 0.21)

  P(X < 0.29) = P( < ) = P(Z < -1) = 1 - P(Z 1)

                                                       = 1 - 0.84134 = 0.15866

  P(X 0.21) = P( ) = P(Z -9) = 1 - P(Z < 9)

Now, for finding the P(Z < 9) in the z table we can observe that the maximum x value which the z table represent is 4.40 with an area of 0.99999. So, we can assume the P(Z < 9) to be 0.99999.                                             

So, P(X 0.21) = 1 - P(Z < 9) = 1 - 0.99999 = 0.00001

Therefore, P(0.21 < X < 0.29) = 0.15866 - 0.00001 = 0.15865

Hence, probability that the diameter of a randomly selected pencil will between 0.21 and 0.29 inches is 0.15865.

the answer is x=13/2 :)