The weights of steers in a herd are distributed normally. The standard deviation is 200lbs and the mean steer weight is 1000 lbs. Find the probability that the weight of a randomly selected steer is between 639 and 1420lbs. Round your answer to four decimal places.

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

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:

Find the probability that the weight of a randomly selected steer is between 639 and 1420lbs.

This is the pvalue of Z when X = 1420 subtracted by the pvalue of Z when X = 639. So

X = 1420

has a pvalue of 0.9821

X = 639

has a pvalue of 0.0355

0.9821 - 0.0355 = 0.9466

0.9466 = 94.66% probability that the weight of a randomly selected steer is between 639 and 1420lbs.

The advertiser failed to reject the null hypotenuse so this would mean there is not sufficient evidence at the 0.02 level of significance to say that the percentage of people who own a Rolls Royce is higher than 47%