Question

question 6
it is also known that customers will spend an average of $178 on additional maintenance. the population
standard deviation of the expenses is $50. if a simple random sample of 100 customers is taken.
what is the probability that the sample mean will be between $166.75 and $170.50?
what is the probability that the sample mean will be less than $189.50?
four decimal places
places

Explanation:

Step1: Calculate the standard error

The formula for the standard error \(SE=\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 50\) and \(n = 100\).
\(SE=\frac{50}{\sqrt{100}}=\frac{50}{10}=5\)

Step2: Calculate the z - scores

For \(x_1 = 166.75\), \(z_1=\frac{x_1-\mu}{SE}=\frac{166.75 - 178}{5}=\frac{- 11.25}{5}=-2.25\)
For \(x_2 = 170.50\), \(z_2=\frac{x_2-\mu}{SE}=\frac{170.50 - 178}{5}=\frac{-7.5}{5}=-1.5\)

Step3: Find the probability using the standard normal distribution

\(P(-2.25<Z<-1.5)=\Phi(-1.5)-\Phi(-2.25)\)
From the standard normal table, \(\Phi(-1.5)=0.0668\), \(\Phi(-2.25)=0.0122\)
\(P(-2.25 < Z<-1.5)=0.0668 - 0.0122=0.0546\)

For the second part:

Step1: Calculate the z - score

For \(x = 189.50\), \(z=\frac{x-\mu}{SE}=\frac{189.50 - 178}{5}=\frac{11.5}{5}=2.3\)

Step2: Find the probability using the standard normal distribution

\(P(Z<2.3)=\Phi(2.3)\)
From the standard normal table, \(\Phi(2.3)=0.9893\)

Answer:

The probability that the sample mean is between \(166.75\) and \(170.50\) is \(0.0546\). The probability that the sample mean is less than \(189.50\) is \(0.9893\)