The number of diners at a restaurant each day is recorded and a daily average is calculated every month (assume 30 days in a month). The number of diners each day has a mean of 109 and a standard deviation of 53, but does not necessarily follow a normal distribution. The probability that a daily average over a given month is greater than x is 2.5%. Calculate x. Give your answer to 3 decimal places.

Respuesta :

Answer:  212.88

Step-by-step explanation:

Given : The probability that a daily average over a given month is greater than x = [tex]2.5\%=0.025[/tex]

The probability that corresponds to  0.025 from a Normal distribution table is 1.96.

Mean : [tex]\mu = 109[/tex]

Standard deviation : [tex]\sigma = 53[/tex]

The formula for z-score : -[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

[tex]\Rightarrow\ 1.96=\dfrac{x-109}{53}\\\\\Rightarrow\ x=53\times1.96+109\\\\\Rightarrow\ x=212.88[/tex]

Z scores (converted value in standard normal distribution) can be mapped to probabilities by z tables. The value of x is 212.88 approx.

How to get the z scores?

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

For the given case, let the random variable X tracks the number of dinners at given restaurant. Assuming normal distribution being pertained by X, we get:

[tex]X \sim N(109, 53)[/tex]

The given data shows that:

2.5% of all daily averages records lie bigger than value X = x

or

P(X > x) = 2.5%  0.025

Converting it to standard normal distribution(so that we can use z tables and p values to get the unknown x), we get:

[tex]z = \dfrac{x-\mu}{\sigma} = \dfrac{x - 109}{53}[/tex]

The given probability statement is expressed as:

[tex]P(Z > z) = 2.5\% = 0.025\\P(Z \leq z) = 1 - 0.025 = 0.975[/tex]

Seeing the z tables, we will try to find at what value of z, the p value is obtained near to 0.975

We get z = 1.96.

Thus,

[tex]z = 1.96 = \dfrac{x - 109}{53}\\\\x = 1.96 \times 53} + 109 = 212.88[/tex]

Thus,

The value of x is 212.88 approx.

Learn more about standard normal distributions here:

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