Normal distribution is a symmetric distribution where the single peak is at the mean of the data. The normal distribution curve is bell shaped and the spread of data is controlled by the standard deviation. The 68-95-99.7 rule says that 68 percent of data in a normal distribution comes under one standard deviation, 95 percent comes under two standard deviations and 99.7 percent of data comes under three standard deviations. To get the probability of from a normal distribution, we need to know the mean and the standard deviation of the given data. Then, calculating the z-score and looking up to the z- table gives the needed probability.

Formula to find the z-score is, $z$ = $\frac{X - \mu}{\sigma}$

Formula to find the z-score is, $z$ = $\frac{X - \mu}{\sigma}$

The ages of all students in a class is normally distributed. The 95 percent of total data is between the age 15.6 and 18.4. Find the mean and standard deviation of the given data.

From 68-95-99.7 rule, we know that in normal distribution 95 percent of data comes under 2-standard deviation.

Mean of the data = $\frac{15.6 + 18.4}{2}$ = $17$

From the mean, 17, to one end, 18.4, there are two standard deviations.

Standard deviation = $\frac{18.4 - 17}{2}$ = $0.7$

If mean of a given data for a random value is 81.1 and standard deviation is 4.7, then find the probability of getting a value more than 83.

Standard deviation, $\sigma$ = $4.7$

Mean, Mean $\mu $ = 81.1

Expected value, X = 83

Z-score, $z$ = $\frac{X - \mu}{\sigma}$

$z$ = $\frac{83 - 81.1}{4.7}$ = $0.404255$

Looking up the z-score in the z-table, 0.6700.

Hence, probability $(1 - 0.6700)$ = $0.33$.

The average speed of a car is 65 kmph with a standard deviation of 4. Find the probability that the speed is less than 60 kmph.

Mean $\mu$ = 65

Standard deviation, $\sigma$ = 4

Expected value, X = 4

Z-score, $z$ = $\frac{X - \mu}{\sigma}$

z = $\frac{60 - 65}{4}$ = -$1.25$

Looking up the z-score in the z-table, we get 0.1056.

Hence, probability is 0.1056.

The average score of a statistics test for a class is 85 and standard deviation is 10. Find the probability of a random score falling between 75 and 95.

The probability of score falling between 75 and 95 can be found after finding the respective z-scores.

For X = 75, z = $\frac{75 - 85}{10}$ = -$1.00$

For X = 95, z = $\frac{95 - 85}{10}$ = $2.00$

Probability is, P(-1.00 < z < 2.00) = P(z < 2.00) - P(z < -1.00)

= 0.9772 - 0.1587 = 0.8185.