Question 1170396: how much moeny does a typical family of four spend at a mcdonalds restaurant per visit?the amount is a normally distributed random variable with a mean of $16.40 and a standard deviation of $2.75
a.find the probability that a family of four spends less than $10
b.what is the amount below which only 10%of families of four spend at mcdonalds?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $X$ be the amount of money a typical family of four spends at a McDonald's restaurant per visit. We are given that $X$ is normally distributed with a mean $\mu = 16.40$ and a standard deviation $\sigma = 2.75$.
a. We want to find the probability that a family of four spends less than $10, i.e., P(X < 10)$.
First, we calculate the z-score:
$$z = \frac{x - \mu}{\sigma} = \frac{10 - 16.40}{2.75} = \frac{-6.40}{2.75} \approx -2.327$$
Now, we want to find $P(Z < -2.327)$, where $Z$ is a standard normal random variable. Using a standard normal table or a calculator, we find that
$$P(Z < -2.327) \approx 0.0100$$
Therefore, $P(X < 10) \approx 0.01$.
b. We want to find the amount below which only 10% of families of four spend at McDonald's. In other words, we want to find the value $x$ such that $P(X < x) = 0.10$.
First, we find the z-score corresponding to a cumulative probability of 0.10. Using a standard normal table or a calculator, we find that the z-score is approximately -1.28.
Now, we use the z-score formula to find $x$:
$$z = \frac{x - \mu}{\sigma}$$
$$-1.28 = \frac{x - 16.40}{2.75}$$
$$x - 16.40 = -1.28 \times 2.75$$
$$x - 16.40 = -3.52$$
$$x = 16.40 - 3.52$$
$$x = 12.88$$
Therefore, the amount below which only 10% of families of four spend at McDonald's is approximately $12.88.
Final Answers:
a. The probability that a family of four spends less than $10 is approximately 0.01.
b. The amount below which only 10% of families of four spend at McDonald's is approximately $12.88.
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