normal distribution: The height X of a randomly chosen femal

The height X of a randomly chosen female is a variable having a normal
distribution with mean of 160 cm and standard deviation of 6 cm. What is the
probability that a randomly chosen female will have height between 160 cm and
170 cm? What is the probability that the sample mean of a sample of 20 females
lies between 158cm and 162cm?

i had a go, is this right?

for the first bit

. . .\(\displaystyle Z = \frac{x\, -\, 160}{6}\)

. . .\(\displaystyle P(160\, \leq\, x\, \leq\, 170)\)

. . .\(\displaystyle P(Z\, \leq\, 1.666...7)\, -\, P(Z\, \leq\, 0)\)

. . .\(\displaystyle =\,0.9515\, -\, 0.5000\)

. . .\(\displaystyle =\,0.4515\)

for the second bit

. . .\(\displaystyle X\, \sim\, N(160,\, \frac{36}{20})\)

. . .\(\displaystyle Z\, =\, \frac{x\, -\, 160}{1.8}\)

. . .\(\displaystyle P(158\, \leq\, x\, \leq\, 162)\)

. . .\(\displaystyle P(Z\, \leq\, -1.1...)\, -\, P(Z\, \leq\, 1.1...)\)

. . .\(\displaystyle =\, 0.8665\, -\, 0.1335\)

. . .\(\displaystyle =\,0.733\)

thanks for your help.

For the second problem, the standard deviation is \(\displaystyle \sqrt{36/20}\).

so, is this right for the second bit?
\(\displaystyle X \sim N(160, \sqr{\frac{36}{20}})\)

\(\displaystyle Z = \frac{x - 160}{ \sqr{1.8}}\)

\(\displaystyle P(158 \leq x \leq 162)\)

\(\displaystyle P(Z \leq 1.49) - P(Z \leq -1.49)\)

\(\displaystyle = 0.9319 - 0.1335\)

\(\displaystyle =0.8638\)