Answer:
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
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:

What is the probability that a randomly selected individual will be between 185 and 190 pounds?
This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So
X = 190



has a pvalue of 0.8944
X = 185



has a pvalue of 0.7357
0.8944 - 0.7357 = 0.1587
15.87% probability that a randomly selected individual will be between 185 and 190 pounds
Answer:
The value of the car in very nice condition is $28,500.
Step-by-step explanation:
Consider the provided information.
In 1955 an antique car that originally cost 3,943 is valued at 64,125 if in excellent condition, winch is 2 1/4 times as much as a car in very nice condition.
The mixed fraction can be written as:

Let x is the value of car in very nice condition.



Hence, the value of the car in very nice condition is $28,500.
30% of 250 can be written as 0.3 * 250:
0.3(250) = 75
30% of 250 is 75. Is this what you mean by "compute with percents?"