Confidence Intervals using Normal Distribution with variance known From a sample of $32$ observations from a population with $σ^2$
 $= 24.602$, we obtained $x = 79.47$. Construct a $99$% confidence interval for $\mu$.  
$(a) 79.47±1.4424$
(b) 79.47±1.7186
$(c) 79.47±2.0430$
$(d) 79.47±2.2622$
Why is the bolded (b) the answer? I keep getting (d).
 A: Respectively, answers (A) through (D) correspond to symmetric two-sided $90\%$, $95\%$, $98\%$, and $99\%$ confidence intervals.  The critical values for each are:
(A) $1.64485$, (B) $1.95996$, (C) $2.32635$, (D) $2.57583$,
with standard errors being $\sqrt{24.602/32} \approx 0.87682$.  Thus the respective margins of error are ($\pm$)
(A) $1.44224$, (B) $1.71853$, (C) $2.03979$, (D) $2.25854$.
A: Using R
 n = 32
 sigma = 4.96
 SE = sigma/sqrt(n)
 ME = qnorm(.995)*SE
 x = 79.47
 x+c(-ME,ME)

=($77.21, 81.73$)
Answer D
Is it possible the book made an error??
With some further explanation, I am certain the book made an error
Here is what the book used. They used a 95% confidence interval
 n = 32
 sigma = 4.96
 SE = sigma/sqrt(n)
 ME = qnorm(.975)*SE
 x = 79.47
 x+c(-ME,ME)

= ($77.75, 81.19$)
You were right, the answer is in fact D. Hopefully the code isn't overwhelming, I could have done the same process by hand, but the book used z = 1.96 for a 95% confidence interval, rather than using 2.58 for a 99% confidence interval
