housing prices in a certain neighborhood average at 97.86 per square foot. If one house in this neighborhood is 1400 square feet
2 answers:
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Answer:
(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.
(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.
(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.
(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size
(e) The population mean may be larger than 75 minutes between irruption.
Step-by-step explanation:
We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.
(a) Let X = <u><em>the interval of time between the eruption</em></u>
So, X ~ Normal()
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = population mean time between irruption = 75 minutes
= standard deviation = 20 minutes
Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)
P(X > 84 min) = P( >
) = P(Z > 0.45) = 1 - P(Z
0.45)
= 1 - 0.6736 = <u>0.3264</u>
The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.
(b) Let = <u><em>sample time intervals between the eruption</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean time between irruption = 75 minutes
= standard deviation = 20 minutes
n = sample of time intervals = 13
Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( > 84 min)
P( > 84 min) = P(
>
) = P(Z > 1.62) = 1 - P(Z
1.62)
= 1 - 0.9474 = <u>0.0526</u>
The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.
(c) Let = <u><em>sample time intervals between the eruption</em></u>
The z-score probability distribution for the sample mean is given by;
Z = ~ N(0,1)
where, = population mean time between irruption = 75 minutes
= standard deviation = 20 minutes
n = sample of time intervals = 20
Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( > 84 min)
P( > 84 min) = P(
>
) = P(Z > 2.01) = 1 - P(Z
2.01)
= 1 - 0.9778 = <u>0.0222</u>
The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.
(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.
(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.
The surface area of the exhibit is 1,638 cubic inches.
Step-by-step explanation:
Step 1:
If three prisms are out on top of each other we get a shape as follows;
The length of the base inches,
The width of the base inches, and
The height of the exhibit inches.
To find the surface area of the total exhibit we consider there to be only one shape which the above dimension.
Step 2:
There are 6 faces to the exhibit.
The surface area of a rectangular prism is given by;
By substituting the values, we get
The surface area of the exhibit is 1,638 cubic inches.