Answer:
Std dev = 0.50
Step-by-step 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
[tex]P(49.74<X<54.86)= 0.80[/tex]
For 49.74 we have x-mean = [tex]49.74-52.30 = -2.56[/tex]
For 54.86 we have x-mean = [tex]54.86-52.30 = 2.56[/tex]
Hence we have
[tex]P(|x-52.3|<2.56) = 0.80\\P(Z<2.56\sigma) = 0.80[/tex]
[tex]2.56\sigma = 1.28\\\sigma = 0.50[/tex]
he standard deviation of the amount of money spent by high school students monthly=0.50