Question

The daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6. What is the probability that a randomly selected bill will be at least $39.10

Answer 1

Answer:

Probability that a randomly selected bill will be at least $39.10 is 0.03216.

Step-by-step explanation:

We are given that the daily dinner bills in a local restaurant are normally distributed with a mean of $28 and a standard deviation of $6.

Let X = daily dinner bills in a local restaurant

So, X ~ N($\mu=28,\sigma^{2} =6^{2}$)

The z-score probability distribution for normal distribution is given by;

               Z = $\frac{ X -\mu}{\sigma}$  ~ N(0,1)

where, $\mu$ = mean amount = $28

            $\sigma$ = standard deviation = $6

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

So, the probability that a randomly selected bill will be at least $39.10 is given by = P(X $\geq$ $39.10)

  P(X $\geq$ $39.10) = P( $\frac{ X -\mu}{\sigma}$ $\geq$ $\frac{ 39.10-28}{6}$ ) = P(Z $\geq$ 1.85) = 1 - P(Z < 1.85)

                                                           = 1 - 0.96784 = 0.03216

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.85 in the z table which has an area of 0.96784.

Hence, the probability that a randomly selected bill will be at least $39.10 is 0.03216.