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

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

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 =
`( 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(
`( X -\mu)/(\sigma)`
`\geq`
`( 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.