We can choose the average number of ounces (µ) a bottling machine puts into each bottle. The actual amount put into each bottle has a normal distribution with µ and a standard deviation of 0.3. If the bottles are 8 ounces, what value of µ would give us a 1% probability of putting more than 8 ounces in a bottle (overflowing) normal distribution

Step-by-step explanation:

Problems of normally distributed samples are 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:

If the bottles are 8 ounces, what value of µ would give us a 1% probability of putting more than 8 ounces in a bottle (overflowing) normal distribution

This means that Z when X = 8 has a pvalue of 1-0.01 = 0.99. So when X = 8, Z = 2.33.

We want to find

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

overpriced

step-by-step explanation: