Answer:
The probability that the average final exam grade of this sample is between 77 and 82
P(77≤ x⁻≤ 82 = 0.8315 or 83%
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given random sample size 'n' = 40
Mean of the normal distribution = 81
Standard deviation of normal distribution = 6.6
Let x₁⁻ = 77
![Z_{1} = \frac{x_{1} -mean }{\frac{S.D}{\sqrt{n} } } = \frac{77-81}{\frac{6.6}{\sqrt{40} } }](https://tex.z-dn.net/?f=Z_%7B1%7D%20%3D%20%5Cfrac%7Bx_%7B1%7D%20-mean%20%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%3D%20%5Cfrac%7B77-81%7D%7B%5Cfrac%7B6.6%7D%7B%5Csqrt%7B40%7D%20%7D%20%7D)
Z₁ = -3.83
Let x₂⁻ = 77
![Z_{2} = \frac{x^{-} _{2} -mean }{\frac{S.D}{\sqrt{n} } } = \frac{82-81}{\frac{6.6}{\sqrt{40} } }](https://tex.z-dn.net/?f=Z_%7B2%7D%20%3D%20%5Cfrac%7Bx%5E%7B-%7D%20_%7B2%7D%20-mean%20%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%3D%20%5Cfrac%7B82-81%7D%7B%5Cfrac%7B6.6%7D%7B%5Csqrt%7B40%7D%20%7D%20%7D)
Z₂ = 0.958
<em>The probability that the average final exam grade of this sample is between 77 and 82</em>
P(77≤ x≤ 82) = P( -3.83 ≤x≤0.958)
= A( 0.958) + A(3.83)
= 0.3315 + 0.4995
= 0.8315
<em>The probability that the average final exam grade of this sample is between 77 and 82 = 0.8315 or 83%</em>
Cross multiply: 60%/100% =21/x
X is the variable we are trying to find.
100*21=2,100
2,100/60
X =35
All together in the choir there are 35 members.
I hope this helps!
Answer:
16 i think
Step-by-step explanation:
could i have brain liest tho i 4/5
Answer:
1 is 21 cm
2 is 23 ft
3 is 19.3 in
4 is 38.6 m
5 is 37.1 yd
6 is 21.6 km
Step-by-step explanation:
just add all thim together 1 by 1