The monthly utility bills in a city are normally distributed with a mean of $121 and a standard deviation of $23. find the probability that a randomly selected utility bill is between $110 and $130.

Step-by-step explanation:

Applying the formula for normal distribution,

z = (x - u)/s

Where u = mean

s = standard deviation

x = the monthly utility bill in dollars

From the information given,

s = 23

u = 121

The probability that a randomly selected utility bill is between $110 and $130 is expressed as

P(110 lesser than or equal to x lesser than or equal to 130)

For 110

z1 = (110 - 121)/23 = - 11/23

z1 = - 0.4783

Looking at the normal distribution table,

The corresponding z score is 0.3192

For 130

z2 = (130 - 121)/23 = 9/23

z2 = 0.391

Looking at the normal distribution table,

The corresponding z score is 0.65173

P(110 lesser than or equal to x lesser than or equal to 130) = 0.65173 - 0.3192 = 0.33253

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

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