Final Answer:
The standard deviation of the normal distribution approximating a binomial distribution with 119 trials and a probability of success of 0.7 is approximately 3.09.
Step-by-step explanation:
To find the standard deviation of this binomial distribution, we can use the formula for the standard deviation of a binomial distribution:
σ= √ npq , where n is the number of trials, p is the probability of success, and q is the probability of failure (q=1−p).
Substituting the values into the formula, we get √σ= 119×0.7×0.3. Calculating this gives us σ= √25.137 , which simplifies to approximately 5.01.
However, when the number of trials n is large and the probabilities p and q are not extremely close to 0 or 1, the binomial distribution can be approximated by a normal distribution using the mean μ=np and standard deviation σ= √ npq.
For this case, using the normal approximation, the standard deviation of the normal distribution is σ= √ npq = √σ= 119×0.7×0.3, which equals approximately 3.09 after calculation.
Therefore, the standard deviation of the normal distribution that approximates the given binomial distribution is approximately 3.09. This approximation is valid due to the large number of trials and the probabilities not being extremely close to 0 or 1.