Question

The monthly utility bills in a city are normally distributed, with a mean of $100 and a standard deviation of $13. Find the probability that a randomly selected (a) less than $70. (b) between $85 and $100, and (c) more than $110. 

(a) The probability that a randomly selected utility bill is less than $70 is _______ 

Answer 1

The monthly utility bills in a city are normally​ distributed, with a mean of $100 and a standard deviation of $13. Find the probability that a randomly selected utility bill is​:

We have,

X $\sim$ N($\mu$ , $\sigma$ ²)

Mean = $\mu$ = $100

Standard deviation = $\sigma$ = $13

Z = (X-$\mu$) / $\sigma$ = (X-100) / 13

(a) less than ​$70

P(X < 70) = P((X-$\mu$) / $\sigma$ < (70-100) / 13​​​​​​​)

P(X < 70) = P(Z  < -30 / 13​​​​​​​)

P(X < 70) = P(Z  < -2.31​​​​​​​)

P(X < 70) = 0.01044

P(X < 70) = 0.0104

​(b) between $85 and $100

P(85 < X < 100) = P((85-100) / 13 < (X-$\mu$) / $\sigma$ < (100-100) / 13​​​​​​​)

P(85 < X < 100) = P(-15 / 13 < Z < 0 / 13​​​​​​​)

P(85 < X < 100) = P(-1.15 < Z < 0​​​​​​​)

P(85 < X < 100) = P(Z < 0​​​​​​​) - P(Z < -1.15)

P(85 < X < 100) = 0.50000 - 0.12507

P(85 < X < 100) = 0.37493

P(85 < X < 100) = 0.3749

(c) more than ​$110

P(X > 110) = P((X-$\mu$) / $\sigma$ > (110-100) / 13​​​​​​​)

P(X > 110) = P(Z  > 10 / 13​​​​​​​)

P(X > 110) = P(Z  > 0.77​​​​​​​)

P(X > 110) = 1-P(Z  < 0.77​​​​​​​)

P(X > 110) = 1-0.77935

P(X > 110) = 0.22065

P(X > 110) = 0.2207