Question

1. a tire manufacturer claims its tires last an average of 60,000 miles with a standard deviation of 5,000 miles. if a consumer agency randomly selects 50 tires from the brand, what is the probability the sample mean is less than 58,500 miles? 8. sat math scores are approximately normally distributed with a mean score of 520 and a standard deviation of 115. a tutoring company collects the sat math scores of a random sample of 36 students who completed its program. what is the probability the average sat score of the sample exceeds 550?

Explanation:

Step1: Calculate the z - score for problem 7

The formula for the z - score of a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\).
For problem 7, \(\mu = 60000\), \(\sigma=5000\), \(n = 50\), \(\bar{x}=58500\).
\(z=\frac{58500 - 60000}{\frac{5000}{\sqrt{50}}}=\frac{- 1500}{\frac{5000}{7.071}}\approx\frac{-1500}{707.1}\approx - 2.12\)

Step2: Find the probability for problem 7

Using the standard normal distribution table, \(P(Z\lt - 2.12)\).
Looking up the value in the standard - normal table, \(P(Z\lt - 2.12)=0.0170\)

Step3: Calculate the z - score for problem 8

For problem 8, \(\mu = 520\), \(\sigma = 115\), \(n = 36\), \(\bar{x}=550\)
\(z=\frac{550 - 520}{\frac{115}{\sqrt{36}}}=\frac{30}{\frac{115}{6}}=\frac{30\times6}{115}=\frac{180}{115}\approx1.57\)

Step4: Find the probability for problem 8

\(P(\bar{X}>550)=1 - P(\bar{X}\leq550)\)
Since \(P(\bar{X}\leq550)\) corresponds to \(z\approx1.57\), and from the standard - normal table \(P(Z\leq1.57) = 0.9418\)
\(P(\bar{X}>550)=1 - 0.9418=0.0582\)

Answer:

For problem 7, the probability is \(0.0170\).
For problem 8, the probability is \(0.0582\).