Answer:
The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For the population, we have that:
Mean = 15
Standard deviaiton = 12
Sample of 30
By the Central Limit Theorem
Mean 15
Standard deviation 
Approximately normal
The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.
Answer:
The value of a₁₀ is -1352
Step-by-step explanation:
a₂ = -8
a₅ = -512
Now,
a₂ = -8 can be written as
a + d = -8 ...(1) and
a₅ = -512 can be written as
a + 4d = -512 ...(2)
Now, from equation (2) we get,
a + 4d = - 512
a + d + 3d = - 512
(-8) + 3d = - 512 (.°. <u>a + d = </u><u>-8</u><u>)</u>
3d = - 512 + 8
3d = - 504
d = - 504 ÷ 3
d = - 168
Now, for the value of a put the value of d = -168 in equation (1)
a + d = -8
a + (-168) = -8
a - 168 = -8
a = 168 - 8
a = 160
Now, For a₁₀
a₁₀ = a + 9d
a₁₀ = 160 + 9(-168)
a₁₀ = 160 - 1512
a₁₀ = -1352
Thus, The value of a₁₀ is -1352
<u>-TheUnknownScientist</u>