Question

On average, 63% of guests at a restaurant order the chef's steak special. The chef is

expecting 220 tomorrow evening. How many of those meals should he plan on
serving in order to be 95% sure (2 standard deviations above the mean) of having
enough steaks on hand to meet customer demand?
Note:
I- Round any internediate numbers to
2- Round your final answer to theneargsunteger.

Answer 1

To be 95% sure of having enough steaks to meet customer demand, the chef needs to plan on serving 63 steak meals.

Step-by-step explanation:

To be 95% sure of having enough steaks to meet customer demand, the chef needs to plan on serving 2 standard deviations above the mean. Since the mean percentage of guests ordering the chef's steak special is 63% and the standard deviation is not given in the question, we need to calculate it using the information given. Assuming a normal distribution, we have:

Mean = 63%

Sample Size = 220

Standard Deviation = √(p(1-p)/n) = √[(0.63)(0.37)/220] ≈ 0.028

To find 2 standard deviations above the mean, we multiply the standard deviation by 2 and add it to the mean:

2 Standard Deviations = 2 * 0.028 ≈ 0.056

Steaks to Serve = Mean + 2 Standard Deviations = 63% + 0.056 = 63.056%

Round the final answer to the nearest whole number: 63 steak meals.