Answer:
The probability that the temperature on a given day in February is 22º or lower is 16.7 %
Step-by-step explanation:
Given:
Average Temperature, ![\mu=30.2^{\circ}](https://tex.z-dn.net/?f=%5Cmu%3D30.2%5E%7B%5Ccirc%7D)
Standard Deviation of Temperature, [tex\sigma=8.5^{\circ}[/tex]
We have to find: Probability that the temperature is 22° or lower
.i.e., P[ X ≤ 22 ]
Let X be the Temperature on a given day in February.
We use Z-score to calculate required probability.
First we convert X into Z using this relationship,
![Z=\frac{X-\mu}{\sigma}=\frac{22-30.2}{8.5}=-0.965](https://tex.z-dn.net/?f=Z%3D%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3D%5Cfrac%7B22-30.2%7D%7B8.5%7D%3D-0.965)
Now, Probability [ Z ≤ -0.965 ] = 0.167272 = 16.7 %
Therefore, The probability that the temperature on a given day in February is 22º or lower is 16.7 %
Answer:
( 11.7 , 12.7 )
Step-by-step explanation:
The provided information is:
Sample mean = ![\bar{X} = 12.2](https://tex.z-dn.net/?f=%5Cbar%7BX%7D%20%3D%2012.2)
Standard deviation = ![\sigma=1.6](https://tex.z-dn.net/?f=%5Csigma%3D1.6)
Sample size = n = 49
Significance level = ![\alpha=0.05](https://tex.z-dn.net/?f=%5Calpha%3D0.05)
Thus, the 95% confidence interval is:
![\bar{X}\pm Z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\=12.2\pm Z_{0.05/2}\frac{1.6}{\sqrt{49}}\\=12.2\pm 1.96\times0.2286\\=12.2\pm 0.4480\\=(11.7.12.7)](https://tex.z-dn.net/?f=%5Cbar%7BX%7D%5Cpm%20Z_%7B%5Calpha%2F2%7D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%5C%5C%3D12.2%5Cpm%20Z_%7B0.05%2F2%7D%5Cfrac%7B1.6%7D%7B%5Csqrt%7B49%7D%7D%5C%5C%3D12.2%5Cpm%201.96%5Ctimes0.2286%5C%5C%3D12.2%5Cpm%200.4480%5C%5C%3D%2811.7.12.7%29)
Yes.
Why?
They worked hard to get that answer by typing things and typing requires work.
Which means credit should be earned
Thank you for your time
165 is answer we also use a formula
5 ( n/2(2a + ( n-1)d)
in that n=6 ,a=3 , d= 1