Question
one year consumers spent an average of $22 on a meal at a restaurant. assume that...
one year consumers spent an average of $22 on a meal at a restaurant. assume that the amount spent on a restaurant meal is normally distributed and that the standard deviation is $3. what is the probability that a randomly selected person spent more than $24. probably that a randomly selected person spends between $5 and $15
Homework Answers
Answer #1
A) P(X > 24)
= P((X - 
)/
> (24 -
)/)
= P(Z > (24 - 22)/3)
= P(Z > 0.67)
= 1 - P(Z < 0.67)
= 1 - 0.7486
= 0.2514
b) P(5 < X < 15)
= P((5 - )/ < (X -
)/ < (15 -
)/)
= P((5 - 22)/3 < Z < (15 - 22)/3)
= P(-5.67 < Z < -2.33)
= P(Z < -2.33) - P(Z < -5.67)
= 0.0099 - 0
= 0.0099
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