X=sleeping hours
"X\\sim N(7,0.5)"
this can be convert to standard normal distribution by using,
"z=\\frac{x-\\bar{x}}{\\sigma}=\\frac{x-7}{0.5}\\\\"
a)
"P(X>9)=P(Z>\\frac{9-7}{0.5})\\\\\nP(X>9)=P(Z>4)=0\\\\"
probability that a randomly selected person sleeps more than 9 hours = 0
b)
"P(X\\le5)=P(Z\\le\\frac{5-7}{0.5})\\\\\nP(X\\le9)=P(Z\\le-4)=0\\\\"
probability that a randomly selected person sleeps 5 hours or less = 0
c) point probability of a normal distribution is 0.
Therefore,probability that a randomly selected person sleeps exactly 10 hours = 0
d) since the mean 7 hours,
probability that a randomly selected person sleeps more than 7 hours = 0.5
e)
"P(X>9)=0 \\ and\\ P(X>7)=0.5"
therefore,
"P(7<X<9)=0.5"
percentage of the population gets between 7 and 9 hours of sleep=50%
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