Denote by mu the mean of X. Given
P(X>5.52)=0.305
i.e.,
P((X-mu)/4> (5.52-mu)/4)=0.305
So, letting Z=(X-mu)/4. Then Z~N(0,1). Hence, the above equality is
P(Z>(5.52-mu)/4)=0.305. So,
(5.52-mu)/4=0.51
So,
mu=5.52-4*0.51=3.48
So, the best answer is 3.5
Approx 0.0027
30 percent.
The mean is 0.9592
a mean of 1 and any standard deviation
probability is 43.3%
It is the Standard normal variable.
with mean of and standard deviation of 1.
Approx 0.0027
with mean and standard deviation . Once standardized, , the test statistic follows Standard Normal Probability Distribution.
30 percent.
3
The mean is 0.9592
a mean of 1 and any standard deviation
It is 0.5
I have included two links. A normal random variable is a random variable whose associated probability distribution is the normal probability distribution. By definition, a random variable has to have an associated distribution. The normal distribution (probability density function) is defined by a mathematical formula with a mean and standard deviation as parameters. The normal distribution is ofter called a bell-shaped curve, because of its symmetrical shape. It is not the only symmetrical distribution. The two links should provide more information beyond this simple definition.
probability is 43.3%
a is true.