You have a binomial distribution with large number of trials n = 750 and 
individual probability of success p = 0.14.


You want to approximate it by the normal distribution.
You should use the mean value m = p*n = 0.14*750 = 105 and standard deviation 

    SD =  =  = 9.5026.


Also, you should use the continuity correction factor.


About approximation of the binomial distribution by normal distribution and continuity correction factor 
see your textbook and/or these Internet sources

https://www.statisticshowto.com/probability-and-statistics/binomial-theorem/normal-approximation-to-the-binomial/

https://online.stat.psu.edu/stat414/lesson/28/28.1

https://stats.libretexts.org/Courses/Las_Positas_College/Math_40%3A_Statistics_and_Probability/06%3A_Continuous_Random_Variables_and_the_Normal_Distribution/6.04%3A_Normal_Approximation_to_the_Binomial_Distribution


    For calculations, you may use your calculator (function normcdf), or Excel spreadsheet (function NORMDIST);
    or online calculator https://onlinestatbook.com/2/calculators/normal_dist.html


(a)  In this case, you should calculate the area under the normal curve between 105.5 and 106.5 
                                                                  (using the correction factor)

     P(x = 106) = normcdf(105.5, 106.5, 105, 9.5026)= 0.0417 (rounded)



(b)  In this case, you should calculate the area under the normal curve on the right from 105.5
                                                                  (using the correction factor)

     P(x >= 106) = normcdf(105.5, 9999, 105, 9.5026) = 0.479  (rounded).



(c)  In this case, you should calculate the area under the normal curve on the left from 105.5
                                                                  (using the correction factor)

     P(x < 106) = normcdf(-9999, 105.5, 105, 9.5026) = 0.521  (rounded).   

                               Notice that it is the complement to the value of probability in (b)



(d)  In this case, you should calculate the area under the normal curve between 103.5 and 106.5
                                                                  (using the correction factor)

     P(104 <= x <= 106) = normcdf(103.5, 106.5, 105, 9.5026) = 0.125  (rounded).