A survey of college students reported that they spend an average of $9.50 a day on...

A survey of college students reported that they spend an average of $9.50 a day on dinner with a standard deviation of $3. What is the probability that 100 randomly selected college students will spend less than $10.00 on average for dinner? Round your answer to 4 decimal places.

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Expert Solution

Solution :

Given that ,

mean = = 9.50

standard deviation = = 3

n = 100

= 9.50

= / n = 3/ 100=0.3

P( <10 ) = P[( - ) / < (10-9.50) /0.3 ]

= P(z <1.67 )

Using z table

= 0.9525

probability= 0.9525

orchestra answered 1 year ago

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