Suppose commute times in a large city are normally distributed and that 62.50% of commuters in this city take more than 21 minutes to commute one-way. if the standard deviation of such commutes is 6.2 minutes, what is the mean commute?

Respuesta :

if 62.5% take more than 21 minutes 
37.5% take less than 21 minutes 

convert .375 to a zscore using the normal distribution chart in your book 

P(.375) = z -0.318639364 

(21 - x) / 6.2 = -0.318639364 

(21 - x) = -1.9755640568 

22.9755640568 = x 

mean commute is 22.98 minutes

The commuters in the city follows a normal distribution.

The mean commute in the large city is 22.984

The given parameters are:

[tex]\mathbf{x = 21}[/tex] -- number of minutes

[tex]\mathbf{\sigma =6.2}[/tex] -- standard deviation

[tex]\mathbf{P(x > 21) =62.50\%}[/tex]

Using the complement rule

[tex]\mathbf{P(x \le 21) =1 - 62.50\%}[/tex]

[tex]\mathbf{P(x \le 21) =0.375}[/tex]

From z-table of probabilities, the z-score, when p value = 0.375 is:

[tex]\mathbf{z = -0.32}[/tex]

The z-score of a distribution is:

[tex]\mathbf{z = \frac{x - \mu}{\sigma}}[/tex]

Substitute known values

[tex]\mathbf{-0.32 = \frac{21 - \mu}{6.2}}[/tex]

Multiply both sides by 6.2

[tex]\mathbf{-1.984 = 21 - \mu}[/tex]

Collect like terms

[tex]\mathbf{\mu= 21 +1.984 }[/tex]

[tex]\mathbf{\mu= 22.984 }[/tex]

Hence, the mean commute is 22.984

Read more about normal probabilities at:

https://brainly.com/question/6476990