Respuesta :
Answer:
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([tex]\mu = 0.30,\sigma^{2} = 0.01^{2}[/tex])
The z-score probability distribution is given by ;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean diameter = 0.30
[tex]\sigma[/tex] = 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 [tex]\leq[/tex] 0.21)
P(X < 0.29) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{0.29-0.30}{0.01}[/tex] ) = P(Z < -1) = 1 - P(Z [tex]\leq[/tex] 1)
= 1 - 0.84134 = 0.15866
P(X [tex]\leq[/tex] 0.21) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{0.21-0.30}{0.01}[/tex] ) = P(Z [tex]\leq[/tex] -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 [tex]\leq[/tex] 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.
