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.
<em>Let X = daily dinner bills in a local restaurant</em>
So, X ~ N()
The z-score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = mean amount = $28
= 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 $39.10)
P(X $39.10) = P(
) = P(Z
1.85) = 1 - P(Z < 1.85)
= 1 - 0.96784 = 0.03216
<em>Now, in the z table the P(Z </em><em> 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.</em>
Hence, the probability that a randomly selected bill will be at least $39.10 is 0.03216.
