(ii) It was found that the mean length of 200 diodes (LED) produced by a company was 20.04 mm with a standard deviation of 0.02mm. Find the probability that a diode
selected at random would have a length less than 20.01mm

6.68% probability that a diode selected at random would have a length less than 20.01mm

Step-by-step explanation:

When the distribution is normal, we use 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 question:

Find the probability that a diode selected at random would have a length less than 20.01mm

This is the pvalue of Z when X = 20.01. So

has a pvalue of 0.0668

6.68% probability that a diode selected at random would have a length less than 20.01mm

b.an increase in a causes b to decrease

0.85 or 85%

step-by-step explanation:

1) divide 23 by 27

2) round the decimal to the nearest 10

3) so your answer would be 0.85 or 85%