Normail approximation to binomial distribution

Suppose X1+X2+...+X80 follows a binomial distribution with n = 80, p = 0.74, q = 0.26, mean = 59.2, variance = 15.392

then we may say that X1 + X2 + ... + X80 follows a normal distribution(approximately) with mean = 59.2, variance = 15.392

Yi is another random variable such that Yi = 7Xi - 1 for i=1, 2, ..., 80

Y1 + Y2 + ... + Y80 = 7(X1 + X2 + ... + X80) - 80

so Mean(Y1 + Y2 + ... Y80) = 7Mean(X1+X2+...+X80) - 80 = 7*59.2 - 80 = 334.4

Var(Y1 + Y2 + ... + Y80) = 49Var(X1+X2+...+X80) = 49 * 15.392 = 754.208

so we may also say that Y1+Y2+...+Y80 follows a normal distribution with mean = 334.4, variance = 754.208

Is this correct up to this point?

Now suppose we want to calculate:

probability(Y1+Y2+...+Y80 >= 340)

=P( Z >= (339.5- 334.4)/sqrt(754.208) )

=P( Z >= 0.1857 )

On the other hand

probability(Y1+Y2+...+Y80 >= 340)

=P( 7(X1+X2+...+X80)-80 >= 340 )

=P( X1+X2+...+X80 >= 60 )

=P( Z >= (59.5-59.2)/sqrt(15.392) )

=P( Z >= 0.0765 )

which is completely different.

It seems the methods do not give consistent answers.

Now when we want to calculate

probability(Y1+Y2+...+Y80 >= 300)

=P( Z >= (299.5 - 334.4)/sqrt(754.208) )

=P( Z >= -1.2708 )

on the other hand

P(Y1+Y2+...+Y80 >= 300)

=P( 7(X1+X2+...+X80)-80 >= 300 )

=P( X1+X2+...+X80 >= 55 )

=P( Z >= (54.5-59.2)/15.392 )

=P( Z >= -1.1980 )

which is quite close to the answer from method 1

So which method is the correct one?

I also don't understand the reasons for such a large discrepancy in the first calculation.

Thanks in advance.