Ann had 45 fliers to post around town. Last week, she posted 1/3 of them. This week, she posted 2/5 of the remaining fliers. How
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Answer:
0.9317
Step-by-step explanation:
Standard deviation of the weights = =100 lbs
Mean weight = u = 1200 lbs
We need to find the probability that the weight(x) of a randomly selected steer is between 1000 lbs and 1369 lbs i.e. P(1000 < x < 1369)
Since, weights follow the normal distribution we can use the z values to find the required weight. For this we have to convert both the values to z score. The formula for z scores is:
1000 converted to z scores is:
1369 converted to z scores is:
So, we have to find the values from z table that lie between -2 to 1.69
P( 1000 < x < 1369 ) = P(-2 < z < 1.69)
P(-2 < z < 1.69) = P(z < 1.69) - P(z < -2)
From the z table:
P(z < 1.69) = 0.9545
P(z < -2) = 0.0228
So,
P(-2 < z < 1.69) = 0.9545 - 0.0228 = 0.9317
Thus,
P( 1000 < x < 1369 ) = 0.9317
From this we can conclude that:
The probability that the weight of a randomly selected steer is between 1000 lbs and 1369 lbs is 0.9317