A plane intersects a prism parallel to the base of the prism. The cross section is a polygon with eight sides. The base of the p
2 answers:
You might be interested in
Answer:
(a) 0.7967
(b) 0.6826
(c) 0.3707
(d) 0.9525
(e) 0.1587
Step-by-step explanation:
The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and variance <em>σ</em>² = 36.
(a)
Compute the value of P (X > 5) as follows:
Thus, the value of P (X > 5) is 0.7967.
(b)
Compute the value of P (4 < X < 16) as follows:
Thus, the value of P (4 < X < 16) is 0.6826.
(c)
Compute the value of P (X < 8) as follows:
Thus, the value of P (X < 8) is 0.3707.
(d)
Compute the value of P (X < 20) as follows:
Thus, the value of P (X < 20) is 0.9525.
(e)
Compute the value of P (X > 16) as follows:
Thus, the value of P (X > 16) is 0.1587.
**Use a <em>z</em>-table for the probabilities.
Explanation:
Quadrants 2 and 3 houses points that have negative x-values, while as Quadrants 1 and 4 houses points that have positive x-values.
If 'your sister' completes the text with negative y-value, the point would be in Q3.
If 'your sister' completes the text with positive y-value, the point would be in Q2.
Answer:
A 90% confidence interval of the true mean is [$119.86, $123.34].
Step-by-step explanation:
We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.
Also, the standard deviation of the population was $6.36.
Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;
P.Q. = ~ N(0,1)
where, = sample mean amount of money spent on textbooks = $121.60
= population standard deviation = $6.36
n = sample of students = 36
= population mean
<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>
<em />
So, 95% confidence interval for the population mean, is ;
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( <
<
) = 0.90
<u>90% confidence interval for</u> = [
,
]
= [ ,
]
= [$119.86, $123.34]
Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].