Let X has normal distribution X~N(42,202), then Z=20X−42~N(0,1),
P(20<X<60)=P(2020−42<20X−42<2060−42)=P(−1.1<Z<0.9)=P(Z<0.9)−P(Z≤−1.1)≈0.8159−0.1357=0.6802.
Approximately 100∗0.6802%=68.02% of games have a point total between 20 and 60.
Answer: 68.02%.
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