Answer:
(1) The first quartile of standard normal distribution is -0.67.
(2) The third quartile of standard normal distribution is 0.67.
(3) The percentage of the observations in the standard normal distribution that are suspected outliers is, 0.74%.
Step-by-step explanation:
A standard normal distribution has mean 0 and variance 1.
(1)
The first quartile of any data is the value below which 25% of the data lie.
That is, P (Z < z) = 0.25.
The value of z is -0.67.
Thus, the first quartile of standard normal distribution is -0.67.
(2)
The third quartile of any data is the value below which 75% of the data lie.
That is, P (Z < z) = 0.75.
The value of z is 0.67.
Thus, the third quartile of standard normal distribution is 0.67.
(3)
In case of a standard normal distribution values lying outside the range
(Q₁ - 1.5 IQR, Q₃ + 1.5 IQR)
are considered as outlier.
Compute the range as follows:
Compute the probability of (Z < -2.68) and P (Z > 2.68) as follows:
The probability of a value being an outlier is,
P (Outlier) = P (Z < -2.68) + P (Z > 2.68)
= 0.0037 + 0.0037
= 0.0074
The percentage of the observations in the standard normal distribution that are suspected outliers is, 0.0074 × 100 = 0.74%.
Step-by-step explanation:
Hope this helped...