Question

According to Runzheimer International, a typical business traveler spends an average of $281 per day in Chicago. This cost includes hotel, meals, car rental, and incidentals. A survey of 70 randomly selected business travelers who have been to Chicago on business recently is taken. For the population mean of $281 per day, what is the probability of getting a sample average of more than $ 275 per day if the population standard deviation is $47?

Answer 1

The probability of getting a sample average will be "0.9454".

Given values are:

Population mean,

• μ = 281

Sample average,

• $\bar{x}$ = 275

Standard deviation,

• σ = 47

Randomly selected business travelers,

• n = 70

Now,

→ $P(\bar{x} > 275)=P(Z > \frac{\bar {x} - \mu}{\frac{\sigma}{\sqrt{n} } })$

→                    $= P(Z > \frac{275-281}{\frac{40}{\sqrt{70} } } )$

→                    $= P(Z > -1.6021)$

→                    $= 0.9454$

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