The diameters of a batch of ball bearings are known to follow a normal distribution with a mean 4.0 in and a standard deviation of 0.15 in. If a ball bearing is chosen randomly, find the probability of realizing the following event: (a) a diameter between 3.8 in and 4.3 in, (b) a diameter smaller than 3 9 in, (c) a diameter larger than 4.2 in

a) 88.54% probability of a diameter between 3.8 in and 4.3 in

b) 25.14% probability of a diameter smaller than 3.9in

c) 90.82% probability of a diameter larger than 4.2 in

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

(a) a diameter between 3.8 in and 4.3 in,

This is the pvalue of Z when X = 4.3 subtracted by the pvalue of Z when X = 3.8.

X = 4.3

has a pvalue of 0.9772

X = 3.8

has a pvalue of 0.0918

0.9772 - 0.0918 = 0.8854

88.54% probability of a diameter between 3.8 in and 4.3 in

(b) a diameter smaller than 3 9 in,

This is the pvalue of Z when X = 3.9. So

has a pvalue of 0.2514

25.14% probability of a diameter smaller than 3.9in

(c) a diameter larger than 4.2 in

This is 1 subtracted by the pvalue of Z when X = 4.2. So

has a pvalue of 0.9082

90.82% probability of a diameter larger than 4.2 in

80x^8−68x^6+16^5−20x^4−4x^3+8x^2

hope i'm right lol