Question

A population is normally distributed with mean 41.2 and standard deviation 4.7. Find the following probabilities. (Round your answers to four decimal places.)

(a) p(41.2 < x < 45.9)

(b) p(39.4 < x < 42.6)

(c) p(x < 50.0)

(d) p(31.8 < x < 50.6)

(e) p(x = 43.8)

(f) p(x > 43.8)

Answer 1

# X ~ N(mean=mu=41.2, standard deviation=sigma=4.7)
mu=41.2
sigma=4.7

# Probabilities computed using R-Software

# (a)
# P(41.2 < X < 45.9)=P(X < 45.9)-P(X < 41.2)
pnorm(45.9,mu,sigma)-pnorm(41.2,mu,sigma)
[1] 0.3413447
# P(41.2 < X < 45.9)=0.3413

z1=(41.2-mu)/sigma
z1
[1] 0
z2=(45.9-mu)/sigma
z2
[1] 1
pnorm(z2)-pnorm(z1)
[1] 0.3413447

# (b)
# P(39.4 < X < 42.6)=P(X < 42.6)-P(X < 39.4)
pnorm(42.6,mu,sigma)-pnorm(39.4,mu,sigma)
[1] 0.2662319
# P(39.4 < X < 42.6)=0.2662

# (c) p(x < 50.0)
pnorm(50,mu,sigma)
[1] 0.9694202
# p(x < 50.0)=0.9694

# (d)
# P(31.8 < X < 50.6)=P(X < 50.6)-P(X < 31.8)
pnorm(50.6,mu,sigma)-pnorm(31.8,mu,sigma)
[1] 0.9544997
# P(31.8 < X < 50.6)=0.9545

# (e) p(x = 43.8)
# Probability at a particular point in continuous distribution is zero. Since normal distribution is continuous distribution P(X=43.8)=0

# (f)
# P(X > 43.8)=1-P(X<43.8)
1-pnorm(43.8,mu,sigma)
[1] 0.2900661
# P(X > 43.8) = 0.2901

### Using Statistical table
One needs to Standardize as:
Z=(X-mu)/sigma
Z will follow Standard Normal Distribution i.e. N(0,1)

So, for example first value to find is P(X < 41.2) . It will be Simplified as:
P(X < 41.2) <=> P([(X-mu)/sigma] > [(41.2-41.2)/4.7])
i.e. P(Z < 0)

In the Standard Normal TABLE, see vertically in the column of Z, value of [0.0] and see horizontally [.00]. Corresponding value will give the required probability. P(Z<0)=0.5

Similarly, P(X<45.9)=P[Z<((45.9-41.2)/4.7)]=P(Z<1)=0.84134.

On subtraction, we get the required probability.

Similarly, other probabilities can be computed after standardizing.

Hope it helps.