In: Statistics and Probability
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.
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