For questions 3 and 4, classify each graph as increasing, decreasing, or constant.
1 answer:
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Answer:
(a) The probability that a randomly selected time interval between eruptions is longer than 82 minutes is 0.3336.
(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 minutes is 0.0582.
(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is 0.0055.
(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.
(e) The population mean must be more than 72, since the probability is so low.
Step-by-step explanation:
We are given that a geyser has a mean time between eruptions of 72 minutes.
Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.
(a) Let X = <u><em>the interval of time between the eruptions</em></u>
So, X ~ N()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean time = 72 minutes
= standard deviation = 23 minutes
Now, the probability that a randomly selected time interval between eruptions is longer than 82 minutes is given by = P(X > 82 min)
P(X > 82 min) = P( >
) = P(Z > 0.43) = 1 - P(Z
0.43)
= 1 - 0.6664 = <u>0.3336</u>
The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.
(b) Let = <u><em>sample mean time between the eruptions</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean time = 72 minutes
= standard deviation = 23 minutes
n = sample of time intervals = 13
Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes is given by = P( > 82 min)
P( > 82 min) = P(
>
) = P(Z > 1.57) = 1 - P(Z
1.57)
= 1 - 0.9418 = <u>0.0582</u>
The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.
(c) Let = <u><em>sample mean time between the eruptions</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean time = 72 minutes
= standard deviation = 23 minutes
n = sample of time intervals = 34
Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is given by = P( > 82 min)
P( > 82 min) = P(
>
) = P(Z > 2.54) = 1 - P(Z
2.54)
= 1 - 0.9945 = <u>0.0055</u>
The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.
(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.
(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 minutes, then we conclude that the population mean must be more than 72, since the probability is so low.
Answer:
A) $192
Step-by-step explanation:
Given:
Yearly expense of Car insurance = $1130
Yearly expense of Vacation = $700
Yearly expense of home repairs = $468
we need to find the monthly allowance of Clark family.
First we find the total annual expense of Clark family.
Total Annual expense is equal to sum of Yearly expense of Car insurance and Yearly expense of Vacation and Yearly expense of home repairs.
Framing in equation form we get;
Total annual expense =
Now we know that;
Number of months in a year = 12 months.
Total Monthly allowance will be equal to Total annual expense divided by Number of months in year.
Framing the equation we get;
Total Monthly allowance =
Rounding to nearest dollar we get;
Total Monthly allowance of Clark family is $192.