A soft drink machine outputs a mean of 2323 ounces per cup. The machine's output is normally distributed with a standard deviation of 22 ounces. You have been asked to calculate the probability of putting less than 2424 ounces in a cup. Using the normal distribution tables or appropriate technology, find the area under the standard normal curve.

Area under the normal curve: 0.6915.

69.15% probability of putting less than 24 ounces in a cup.

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:

You have been asked to calculate the probability of putting less than 24 ounces in a cup.

pvalue of Z when X = 24. So

has a pvalue of 0.6915

Area under the normal curve: 0.6915.

69.15% probability of putting less than 24 ounces in a cup.

1gm = 1000mg so 0.1gm = 1000mg

first, convert the weight to kg

160lb * 1kg/2.2lb = 72.7kg

next, the dose

72.7kg * 100mg/kg = 7270mg

y=mx+b

step-by-step explanation:

m slope, b y-intercept