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

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step-by-step explanation:

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;