It is known that IQ scores have a normal distribution with a mean of 100 and a standard deviation of 15

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a) a sample of 36 students is selected. What's the probability that the sample mean IQ score of these 36 students is between $95$ and $110$?

b) same as a but what if the sample size is $100$ instead?

My work: X~N(100, 225)

$\sigma = 15$

$\mu = 100$

a) $n = 36$

Using the central limit theorem as $30 \leq n$

$P(95 < \bar X < 110) = P(\cfrac{95-100}{15 / \sqrt {36}} < Z < \cfrac{110-100}{15/\sqrt{36}}) = P(-2 < Z < 4) = .9772 $(approximately using normalcdf on my calculator)

b) $n = 100$

$P(95 < \bar X < 110) = P(\cfrac{95-100}{15 / \sqrt {100}} < Z < \cfrac{110-100}{15/\sqrt{100}}) = P(-3.33 < Z < 6.67) = .9996 $(approximately using normalcdf on my calculator)

Now I think I'm right but these probabilities seem way too high and I wanted to double check with this website. Did I make a mistake somewhere and am I using the right formula? Also my book uses a table but z = 6.67 is not on the table... so I assumed that I had to use my calculator. Is there any other way?