Answer:
a. X~N(2,885, 651)
b. 0.086291
c. 0.00058
d. 3213.10 calories
Step-by-step explanation:
a. -A normal distribution is expressed in the form X~N(mean, standard deviation).
-Let X a random variable denoting the number of calories consumed.
-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.
-This distribution is expressed as X~N(2,885, 651)
b. The probability that less than 2000 calories are consumed is calculated using the formula:
![P(X](https://tex.z-dn.net/?f=P%28X%3Cx%29%3DP%28z%3C%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%7D%29)
#substitute the given values in the formula to solve for P:
![P(X](https://tex.z-dn.net/?f=P%28X%3Cx%29%3DP%28z%3C%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%7D%29%5C%5C%5C%5C%3DP%28Z%3C%5Cfrac%7B2000-2885%7D%7B651%7D%29%5C%5C%5C%5C%3DP%28z%3C-1.359%29%5C%5C%5C%5C%5C%5C%3D0.08691)
Hence, the probability of consuming less than 2000 calories is 0.08691
c. The proportion of customers consuming more than 5000 calories is calculated as:
![P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058](https://tex.z-dn.net/?f=P%28X%3Ex%29%3DP%28z%3E%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%7D%29%5C%5C%5C%5C%3DP%28Z%3E%5Cfrac%7B5000-2885%7D%7B651%7D%29%5C%5C%5C%5C%3DP%28z%3E3.2488%29%5C%5C%5C%5C%3D1-0.99942%5C%5C%5C%5C%3D0.00058)
Hence, the proportion of customers consuming over 5000 calories is 0.00058
d. The least amount of calories to get the award is calculated as:
1% is equivalent to a z value of 0.50399.
-We equate this to the formula to solve for the mean consumption:
![0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09](https://tex.z-dn.net/?f=0.01%3DP%28z%3E%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%7D%29%5C%5C%5C%5C%3DP%28z%3E%5Cfrac%7B%5Cbar%20X-2885%7D%7B651%7D%29%5C%5C%5C%5C%5C1%5C%25%3D0.50399%20%5C%5C%5C%5C%5Cfrac%7B%5Cbar%20X-2885%7D%7B651%7D%3D0.50399%20%5C%5C%5C%5C%5Cbar%20X%3D0.50399%5Ctimes%20651%2B2885%5C%5C%5C%5C%3D3213.09)
Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.