Question

In: Statistics and Probability

A small university knows the average amount that its students spend on lunch each day. The...

A small university knows the average amount that its students spend on lunch each day. The amount spent on lunches for the population of 500 students is not highly skewed and has a mean of $8 and a standard deviation of $2. Suppose simple random sample of 49 students is taken, what is the probability that the sample mean for the sample of 49 students will be between $7.50 and $8.50?

Expert Solution

Solution :

Given that ,

mean = = 8

standard deviation = = 2

n = 49

= 8

= / $\sqrt$ n= 2/ $\sqrt$ 49 =0.29

P(7.50< <8.50 ) = P[(7.50-8) / 0.29< ( - ) / < (8.50-8) /0.29 )]

= P( -1.72< Z <1.72 )

= P(Z <1.72 ) - P(Z <-1.72 )

Using z table

=0.9573-0.0427

=0.9146

probability= 0.9146

orchestra answered 1 year ago

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