Question

In a distribution, exactly normal, 7% of the items are under 35 and 89% are under 63. Find the mean and standard deviation of the distribution. [Given A(z = 1.48) = 0.43, A(z = 1.23) = 0.39]

Answer 1

μ ≈ 50.29, σ ≈ 10.33

Answer 2

The problem asks us to find the mean ($\mu$) and standard deviation ($\sigma$) of a normal distribution given two probability statements.

We are given:

1. 7% of the items are under 35, which means $P(X < 35) = 0.07$.
2. 89% of the items are under 63, which means $P(X < 63) = 0.89$.

We use the concept of Z-scores to standardize the normal distribution. The Z-score for a value $X$ is given by $Z = \frac{X - \mu}{\sigma}$. The standard normal distribution has a mean of 0 and a standard deviation of 1. The area under the standard normal curve between $z=0$ and $z$ is denoted by $A(z)$ or $\Phi(z) - 0.5$.

From the first condition, $P(X < 35) = 0.07$. We convert this to a Z-score: $P(Z < z_1) = 0.07$, where $z_1 = \frac{35 - \mu}{\sigma}$. Since $0.07 < 0.5$, the value $X=35$ is to the left of the mean, so $z_1$ must be negative. The area to the left of $z_1$ is 0.07. The area to the left of 0 is 0.5. The area between $z_1$ and 0 is $0.5 - 0.07 = 0.43$. The area between 0 and $|z_1|$ is $0.43$. We are given $A(z = 1.48) = 0.43$, which means the area between 0 and 1.48 is 0.43. Since $z_1$ is negative, we have $|z_1| = 1.48$, so $z_1 = -1.48$. Thus, we have the equation: $\frac{35 - \mu}{\sigma} = -1.48$. $35 - \mu = -1.48\sigma$ (Equation 1)

From the second condition, $P(X < 63) = 0.89$. We convert this to a Z-score: $P(Z < z_2) = 0.89$, where $z_2 = \frac{63 - \mu}{\sigma}$. Since $0.89 > 0.5$, the value $X=63$ is to the right of the mean, so $z_2$ must be positive. The area to the left of $z_2$ is 0.89. The area to the left of 0 is 0.5. The area between 0 and $z_2$ is $0.89 - 0.5 = 0.39$. We are given $A(z = 1.23) = 0.39$, which means the area between 0 and 1.23 is 0.39. Therefore, $z_2 = 1.23$. Thus, we have the equation: $\frac{63 - \mu}{\sigma} = 1.23$. $63 - \mu = 1.23\sigma$ (Equation 2)

Now we have a system of two linear equations with two variables $\mu$ and $\sigma$:

1. $35 - \mu = -1.48\sigma$
2. $63 - \mu = 1.23\sigma$

Subtract Equation 1 from Equation 2: $(63 - \mu) - (35 - \mu) = 1.23\sigma - (-1.48\sigma)$ $63 - \mu - 35 + \mu = 1.23\sigma + 1.48\sigma$ $28 = (1.23 + 1.48)\sigma$ $28 = 2.71\sigma$ $\sigma = \frac{28}{2.71}$

Now substitute the value of $\sigma$ into Equation 2 to find $\mu$: $63 - \mu = 1.23 \times \frac{28}{2.71}$ $\mu = 63 - 1.23 \times \frac{28}{2.71}$ $\mu = 63 - \frac{1.23 \times 28}{2.71}$ $\mu = 63 - \frac{34.44}{2.71}$ To simplify, we can write $\mu = \frac{63 \times 2.71 - 34.44}{2.71} = \frac{170.73 - 34.44}{2.71} = \frac{136.29}{2.71}$.

Calculating the numerical values: $\sigma = \frac{28}{2.71} \approx 10.3321$ $\mu = \frac{136.29}{2.71} \approx 50.2915$

Rounding to two decimal places, we get $\mu \approx 50.29$ and $\sigma \approx 10.33$.

Explanation of the solution:

1. Convert the given probabilities $P(X < x)$ into corresponding Z-scores using the Z-score formula $Z = \frac{X - \mu}{\sigma}$ and the provided area values $A(z)$.
2. $P(X < 35) = 0.07 \implies P(Z < z_1) = 0.07$. Since $0.07 < 0.5$, $z_1$ is negative. Area between $z_1$ and 0 is $0.5 - 0.07 = 0.43$. Given $A(1.48) = 0.43$, so $|z_1| = 1.48$. Thus $z_1 = -1.48$. This gives the equation $\frac{35 - \mu}{\sigma} = -1.48$.
3. $P(X < 63) = 0.89 \implies P(Z < z_2) = 0.89$. Since $0.89 > 0.5$, $z_2$ is positive. Area between 0 and $z_2$ is $0.89 - 0.5 = 0.39$. Given $A(1.23) = 0.39$, so $z_2 = 1.23$. This gives the equation $\frac{63 - \mu}{\sigma} = 1.23$.
4. Solve the system of equations: $35 - \mu = -1.48\sigma$ $63 - \mu = 1.23\sigma$ Subtracting the first from the second gives $28 = 2.71\sigma$, so $\sigma = \frac{28}{2.71} \approx 10.33$.
5. Substitute $\sigma$ back into either equation to find $\mu$. Using the second equation: $63 - \mu = 1.23 \times \frac{28}{2.71} \implies \mu = 63 - \frac{34.44}{2.71} = \frac{136.29}{2.71} \approx 50.29$.