HELP ASAP! AGE PROBLEM!! WILL MARK BRAINLIEST!!!
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Substitute the values of A = 3 and B = -4 to the expression:
Answer:
The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.
Step-by-step explanation:
<h2>Q. This is the question: </h2>The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
<h2>Answer:</h2>This is the question of normal distribution:
First w calculate the value of Z corresponding to X = 60,000 miles
We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles
Now, for Z, we know that:
Z = (x-μ)/σ
Z = (60,000 - 50,000)/5,000
<u>Z = 2</u>
Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.
P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)
P(X > 60,000) = 1 - 0.9772
<u>P(X > 60,000) = 0.0228</u>
