Respuesta :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 1000, \sigma = 200[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1420 - 1000}{200}[/tex]
[tex]Z = 2.1[/tex]
[tex]Z = 2.1[/tex] has a pvalue of 0.9821
X = 639
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{639 - 1000}{200}[/tex]
[tex]Z = -1.805[/tex]
[tex]Z = -1.805[/tex] 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.
Answer:
Probability that the weight of a randomly selected steer is between 639 and 1420 lbs is 0.9470.
Step-by-step explanation:
We are given that the weights of steers in a herd are distributed normally. The standard deviation is 200 lbs and the mean steer weight is 1000 lbs.
Let X = weights of steers in a herd
So, X ~ N([tex]\mu=1000,\sigma^{2} = 200^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean steer = 1000 lbs
[tex]\sigma[/tex] = standard deviation = 200 lbs
So, probability that the weight of a randomly selected steer is between 639 and 1420 lbs is given by = P(639 lbs < X < 1420 lbs)
P(639 lbs < X < 1420 lbs) = P(X < 1420 lbs) - P(X [tex]\leq[/tex] 639 lbs)
P(X < 1420 lbs) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{1420-1000}{200}[/tex] ) = P(Z < 2.10) = 0.98214
P(X [tex]\leq[/tex] 639 lbs) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{639-1000}{200}[/tex] ) = P(Z [tex]\leq[/tex] -1.81) = 1 - P(Z < 1.81)
= 1 - 0.96485 = 0.03515
Therefore, P(639 lbs < X < 1420 lbs) = 0.98214 - 0.03515 = 0.9470
Hence, probability that the weight of a randomly selected steer is between 639 and 1420 lbs is 0.9470.
