Answer:
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1.5, 3.25, 1.66, 3.33, 2.4 in least to greatest
2 answers:
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Answer:
Answer → C) 121°
Step-by-step explanation:
» Let the wanted angle be x
Answer:
a) 0.625; b) 16.879; c) 8.442
Step-by-step explanation:
Since this is a normal distribution, we want the value of the mean, μ: 2.2; and the value of the standard deviation, σ: 6.3.
For the 40th percentile, we look in a z chart. We want to find the value as close to 0.40 as we can get; this is 0.4013, and it corresponds to a z score of z = -0.25.
Our formula for a z score is . Using our values, we have
-0.25 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(-0.25) = X-2.2
-1.575 = X-2.2
Add 2.2 to each side:
-1.575+2.2 = X-2.2+2.2
0.625 = X
For the 99th percentile, the value in the z chart closest to 0.99 is 0.9901, which corresponds to a z score of z = 2.33:
2.33 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(2.33) = X-2.2
14.679 = X-2.2
Add 2.2 to each side:
14.679+2.2 = X-2.2+2.2
16.879 = X
For the IQR, we find the values for the 75th percentile (Q3) and the 25th percentile (Q1). The value in a z chart closest to 0.75 is 0.7486, which corresponds to a z score of z = 0.67:
0.67 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(0.67) = X-2.2
4.221 = X-2.2
Add 2.2 to each side:
4.221+2.2 = X-2.2+2.2
6.421 = X
The value in a z chart closest to 0.25 is 0.2514, which corresponds to a z score of z = -0.67:
-0.67 = (X-2.2)/6.3
Multiply both sides by 6.3:
6.3(-0.67) = X-2.2
-4.221 = X-2.2
Add 2.2 to each side:
-4.221+2.2 = X-2.2+2.2
-2.021 = X
This makes the interquartile range
6.421--2.021 = 8.442
Answer:
(a) Null Hypothesis, :
= $1,150
Alternate Hypothesis, :
$1,150
(b) The test statistic is 3.571.
(c) We conclude that the mean of all account balances is significantly different from $1,150.
Step-by-step explanation:
We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.
<u><em>Let </em></u><u><em> = mean of all account balances</em></u>
(a)So, Null Hypothesis, :
= $1,150 {means that the mean of all account balances is equal to $1,150}
Alternate Hypothesis, :
$1,150 {means that the mean of all account balances is significantly different from $1,150}
The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;
T.S. = ~
where, = sample average balance = $1,200
s = sample standard deviation = $126
n = sample of account balances = 81
(b) So, <u>test statistics</u> = ~
= 3.571
The value of the sample test statistics is 3.571.
(c) <u>Now, P-value of the test statistics is given by the following formula;</u>
P-value = P( > 3.571) = <u>Less than 0.05%</u>
Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.
Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the mean of all account balances is significantly different from $1,150.