Question

In a recent survey of high school students, it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $52.30. Suppose you are told that there is an 80% probability that a randomly-selected student spends somewhere between $49.74 and $54.86. What is the standard deviation of the amount of money spent by high school students monthly?

Answer 1

Std dev = 0.50

Explanation:

Given that in a recent survey of high school students, it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $52.30.

Also given that

`P(49.74<X<54.86)= 0.80`

For 49.74 we have x-mean =
`49.74-52.30 = -2.56`

For 54.86 we have x-mean =
`54.86-52.30 = 2.56`

Hence we have

`P(|x-52.3|<2.56) = 0.80\\P(Z<2.56\sigma) = 0.80`

`2.56\sigma = 1.28\\\sigma = 0.50`

he standard deviation of the amount of money spent by high school students monthly=0.50