Which row in the chart correctly identifies the functions of structures A, B, and C? *
1 answer:
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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.
Answer:
At the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.
Step-by-step explanation:
We are given that the mean gross annual incomes of certified welders are normally distributed with the mean of $20,000 and a standard deviation of $2,000.
The ship building association wishes to find out whether their welders earn more or less than $20,000 annually.
<u><em>Let </em></u><u><em> = mean gross annual incomes of certified welders</em></u>
So, Null Hypothesis, :
= $20,000
Alternate hypothesis, :
$20,000
Here, null hypothesis states that the mean income of welders is equal to $20,000.
On the other hand, alternate hypothesis states that the mean income of welders is not $20,000.
Also, the test statistics that would be used here is <u>One-sample z test</u> <u>statistics</u> as we know about the population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean income
= population standard deviation = $2,000
n = sample size
Now, at the 0.10 level of significance the z table gives critical values of -1.645 and 1.645 for two-tailed test.