Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The average amount of money spent for lunch per person in the college cafeteria is $7.15\$7.15 and the standard deviation is $2.48\$2.48. Suppose that 4848 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 44 decimal places where possible. What is the distribution of XX? XN(,X \sim N(,) What is the distribution of xˉ\bar{x}? xˉN(,\bar{x} \sim N(,) For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $6.77\$6.77 and $7.27\$7.27. For the group of 4848 patrons, find the probability that the average lunch cost is between $6.77\$6.77 and $7.27\$7.27. HINTS: $2.48\$2.4800 $2.48\$2.4811 To find $2.48\$2.4822 use normalcdf$2.48\$2.4833 $2.48\$2.4844 $2.48\$2.4855 When you are looking for the probability of one value the $2.48\$2.4866 Submit

Full solution

Q. The average amount of money spent for lunch per person in the college cafeteria is $7.15\$7.15 and the standard deviation is $2.48\$2.48. Suppose that 4848 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 44 decimal places where possible. What is the distribution of XX? XN(,X \sim N(,) What is the distribution of xˉ\bar{x}? xˉN(,\bar{x} \sim N(,) For a single randomly selected lunch patron, find the probability that this patron's lunch cost is between $6.77\$6.77 and $7.27\$7.27. For the group of 4848 patrons, find the probability that the average lunch cost is between $6.77\$6.77 and $7.27\$7.27. HINTS: $2.48\$2.4800 $2.48\$2.4811 To find $2.48\$2.4822 use normalcdf$2.48\$2.4833 $2.48\$2.4844 $2.48\$2.4855 When you are looking for the probability of one value the $2.48\$2.4866 Submit
  1. Given Data: Given μ=$7.15\mu = \$7.15 and σ=$2.48\sigma = \$2.48. For a single lunch patron, the distribution of XX is XN(μ,σ)X \sim N(\mu, \sigma).
  2. Distribution of XX: So, XN(7.15,2.48)X \sim N(7.15, 2.48).
  3. Distribution of xˉ\bar{x}: For the average of 4848 patrons, the distribution of xˉ\bar{x} is xˉN(μ,σ/n)\bar{x} \sim N(\mu, \sigma/\sqrt{n}).
  4. Calculate σ/n\sigma/\sqrt{n}: Calculate σ/n\sigma/\sqrt{n} where n=48n = 48.\newlineσ/n=2.48/482.48/6.92820.3578\sigma/\sqrt{n} = 2.48/\sqrt{48} \approx 2.48/6.9282 \approx 0.3578.
  5. Probability Calculation: So, xˉN(7.15,0.3578)\bar{x} \sim N(7.15, 0.3578).
  6. Use normalcdf: To find the probability that a single patron's lunch cost is between $6.77\$6.77 and $7.27\$7.27, use P(6.77<X<7.27)P(6.77 < X < 7.27).
  7. Calculate Probability: Use normalcdf for a single value with n=1n = 1.P(6.77<X<7.27)=normalcdf(6.77,7.27,7.15,2.48)P(6.77 < X < 7.27) = \text{normalcdf}(6.77, 7.27, 7.15, 2.48).
  8. Calculate Probability: Use normalcdf for a single value with n=1n = 1.P(6.77<X<7.27)=P(6.77 < X < 7.27) = normalcdf(6.77,7.27,7.15,2.48)(6.77, 7.27, 7.15, 2.48).Calculate the probability for a single patron.P(6.77<X<7.27)P(6.77 < X < 7.27) \approx normalcdf(6.77,7.27,7.15,2.48)0.1574(6.77, 7.27, 7.15, 2.48) \approx 0.1574.

More problems from Identify independent events