Q.The daily sales of a retail shop are normally distributed with a mean of ₹5,000 and a standard deviation of ₹800. Find the probability that sales on a given day exceed ₹6,600. (Area between Z = 0 and Z = 2.00 is 0.4772.)

The normal distribution was first obtained by Abraham De-Moivre as the limiting case of the binomial distribution.

Historically, Abraham De-Moivre derived the normal curve in 1733 as an approximation to the binomial distribution for large $n$. Although Gauss and Laplace later developed and popularised it (it is often called the Gaussian distributi …

Checking each statement against the normal distribution's known properties shows the maximum-height claim is false in general — the correct maximum height is $1/(\sigma\sqrt{2\pi})$, not $1/\sqrt{2\pi}$.

1. The normal distribution's pdf is $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}}$.
2. Statement (a): the points of inflection occur where $f''(x)=0$, which for the normal curve are exactly at $x=\mu\pm\sigma$ — this is a true, standard property.
3. Statement (b): the normal distribution is symmetric about $\mu$, so its skewness is $0$ — true. …

Matching the given pdf to the standard normal form $\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$ gives $\sigma=5$, so $c=\dfrac{1}{5\sqrt{2\pi}}$.

1. Standard normal density: $f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(-\dfrac{(x-\mu)^2}{2\sigma^2}\right)$.
2. Given: $f(x)=c\exp\left(-\dfrac12\cdot\dfrac{(x-100)^2}{25}\right)=c\exp\left(-\dfrac{(x-100)^2}{50}\right)$. …

By the symmetry of the normal curve about the mean, $80$ students score between $45$ and $65$.

1. Mean $=65$. Of $400$ students, $120$ scored above $85$, so $P(X>85)=\dfrac{120}{400}=0.30$.
2. Note $85-65=20$ and $65-45=20$: both $45$ and $85$ are exactly $20$ marks from the mean, on opposite sides.
3. By symmetry of the normal distribution about its mean, the area between the mean and a point at distance $d$ above the mean equals the area between the mean and the point at distance $d$ below the mean: $P(65<X<85)=P(45<X<65)$.
4. $P(65<X<85)=P(X<85)-P(X<65)=(1-0.30)-0.5=0.70-0.50=0.20$.
5. So $P(45<X<65)=0.20$ as well, giving $400\times0.20=80$ students. …