Answer: 0.62%
Step-by-step explanation:
Given : The number of calories in peanut M&Ms have an approximately normal distribution with
[tex]\mu=10\text{ calories per }M\&M[/tex]
Standard deviation : [tex]\sigma=\sqrt{\text{Variance}}=\sqrt{4}=2[/tex]
Let x be the amount of calories in a randomly selected peanut M&Ms.
Z-score : [tex]\dfrac{x-\mu}{\sigma}[/tex]
[tex]z=\dfrac{15-10}{2}=2.5[/tex]
The probability that the peanut M&Ms are more than 15 calories using standardized normal distribution table :
[tex]P(x>15)=P(z>2.5)=1-P(z<2.5)\\\\=1-0.9937903=0.0062097[/tex]
In percent , [tex]0.62097\%\approx0.62\%[/tex]
Hence, the percentage of peanut M&Ms are more than 15 calories is 0.62%.