Question

# Suppose you are interested in measuring the amount of time, on average, it takes you to...

Suppose you are interested in measuring the amount of time, on average, it takes you to make your commute to school. You've estimated that the average time is 39 minutes with a standard deviation of 5.611 minutes. Assuming that your estimated parameters are correct and the commute time is normally distributed, what is the probability that the average commute time of 12 random days is greater than 40.72 minutes?

 1) 0.8598
 2) 0.1441
 3) 0.3796
 4) 0.8559
 5) 0.6204

A customer wants to estimate the average delivery time of a pizza from the local pizza parlor. Over the course of a few months, the customer orders 30 pizzas and records the delivery times. The average delivery time is 25.9 with a standard deviation of 9.97. If the customer estimates the time using a 90% confidence interval, what is the margin of error?

 1) 1.8203
 2) 0.7514
 3) 3.0895
 4) 2.3872
 5) 3.0929

Solution :

Given that,

= 25.9

s = 9.97

n = 30

Degrees of freedom = df = n - 1 = 30 - 1 = 29

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t /2,df = t0.05,29=1.6991

Margin of error = E = t/2,df * (s /n)

= 1.6991 * (9.97 / 30) = 3.0929

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