c. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?
1 answer:
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>
You might be interested in
Scarlet is leaning a 12 foot ladder. When you lean a 12 foot ladder, it is s slanted so we will make 12 our slanted side or hypotenuse. It makes a 80 degree angle after contact with the ground so it will be on the lowest angle on the right. We are asked to find the vertical side or how high up the ground is. We are asked to a angle that is opposite of the given angle which is x .. And we are also given the hypotenuse. We are going to use sin using our SOHCAHTOA
Where there is our given angle.
Let plug it in
Multiply 12 by both sides and we get
Which about 11.8 if rounded to nearest tenth or 11.82 if rounded to nearest hundredth.
IF ROUNDED TO NEAREST TENTH, 11.8
IF ROUNDED TO NEAREST HUNDRETH, 11.82