Solution-:We have been given" population average "(mu)=$238.92population standard devicition (sigma)=$31.38let x= spending in playing video games every year forthen{:[x∼Normal(mu","sigma)],[=>quad∼Normal(238.92","31.38)]:}[a] The probability that a family spends over $290 in playing video games every year.{:[p(x > 290)=1-p(x <= 290)],[=1-p(z <= (290-238.92)/(31.38))quad{[" Because "],[x∼N(238.92","3138)]}],[=1-P(z <= 1.627788)],[=1-phi(1.63)=1-0.9484quad{[" From "],[z"-fable "]}]:}[b] The probabilit ... See the full answer