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The Normal Distribution: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

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Problem 1 of 10

Problem 1 The shaded region

The figure shows two shaded regions under the standard normal curve. Given $\Phi(1.2)\approx0.8849$ and $\Phi(2.4)\approx0.9918$ , find the total shaded probability to four decimal places. Here $\Phi(z)$ is the area to the left of z.

The two shaded regions.

Text description of this figure

A bell-shaped standard normal curve drawn over a horizontal z-axis from -4 to 4, with a labeled tick at every integer and two extra labeled ticks, set lower, at -1.2 and 2.4. The region under the curve to the left of -1.2 is shaded, and so is the region to the right of 2.4; a vertical segment rises from each of those ticks to the curve, and a small arrowhead at each end of the axis shows that both shaded tails continue beyond the window. The vertical axis is labeled Density and carries no numbers. No area is labeled.
1. Hint 1
  Read the boundary of each shaded tail.
2. Hint 2
  Use symmetry for the negative boundary, and subtract the positive cumulative area from one for the right tail.
Answer

$P\approx0.1233$ .

Full solution

The shaded regions are below $-1.2$ and above $2.4$ .
Symmetry gives $\Phi(-1.2)\approx0.1151$ .
The area above $2.4$ is approximately
$1-0.9918=0.0082.$
Since the regions do not overlap, their combined area is
$P\approx0.1151+0.0082.$
Thus the total probability is approximately $0.1233$ .

Answer

$P\approx0.1233$ .

Key idea

Separate shaded tails are added after each is found from a cumulative normal area.
Problem 1: Mark complete
Problem 2 Two coordinate labels

For the same normal distribution, the raw value 20 has z-score $-1$ , and the raw value 35 has z-score 2. Find its standard deviation.
1. Hint 1
  Within this distribution, differences in raw values are the corresponding differences in z-scores multiplied by the standard deviation.
2. Hint 2
  The two z-scores are three standard deviations apart.
Answer

$\sigma=5$ .

Full solution

The raw values differ by $35-20=15$ , and their z-scores differ by $2-(-1)=3$ .
Thus
$3\sigma=15,$
so $\sigma=5$ .
The implied mean is $25$ , which gives z-scores $-1$ and $2$ for the stated values.

Answer

$\sigma=5$ .

Key idea

Within one distribution with positive standard deviation, a raw-value difference equals the z-score difference times the standard deviation.
Problem 2: Mark complete
Problem 3 A given density height

Find all real values of $z$ satisfying $\varphi(z)=\frac{e^{-7}}{\sqrt{2\pi}}$ , where $\varphi$ is the standard normal density.
1. Hint 1
  The density depends on the square of the distance from zero.
2. Hint 2
  Cancel the common positive factor and compare the exponents.
Answer

$z=\pm\sqrt{14}$ .

Full solution

Here
$\varphi(z)=\frac{e^{-z^2/2}}{\sqrt{2\pi}}.$
Canceling the positive factor $\frac1{\sqrt{2\pi}}$ leaves $e^{-z^2/2}=e^{-7}$ , and the exponential takes each value once, so
$\frac{z^2}2=7.$
Thus $z^2=14$ and $z=\pm\sqrt{14}$ , one value on each side of the center.

Answer

$z=\pm\sqrt{14}$ .

Key idea

Each positive height below the standard normal curve's peak is reached at exactly two z-values, symmetric about zero.
Problem 3: Mark complete
Problem 4 Two acceptance ranges

A sensor error X is normal with mean 0 and standard deviation 2. Range A accepts $-2<X<2$ ; range B accepts $0<X<4$ . Find both acceptance probabilities to four decimal places and identify the larger one. Use $\Phi(0)\approx0.5000$ , $\Phi(1)\approx0.8413$ , and $\Phi(2)\approx0.9772$ .
1. Hint 1
  Express each endpoint in standard deviations from the mean.
2. Hint 2
  Each probability is an upper cumulative area minus a lower cumulative area; use symmetry below zero.
Answer

Range A: $0.6826$ ; range B: $0.4772$ ; A is larger. Both probabilities are approximate.

Full solution

Range A corresponds to $-1<Z<1$ , so
$\begin{aligned}P(A)&\approx0.8413-0.1587 \\ P(A)&\approx0.6826.\end{aligned}$
Range B corresponds to $0<Z<2$ , so
$\begin{aligned}P(B)&\approx0.9772-0.5000 \\ P(B)&\approx0.4772.\end{aligned}$
Both raw intervals have width $4$ , but A includes more of the high-density region near the mean.

Answer

Range A: $0.6826$ ; range B: $0.4772$ ; A is larger. Both probabilities are approximate.

Key idea

Equal-width intervals can have different probabilities because density is higher near the normal mean.
Problem 4: Mark complete
Problem 5 A range of settings

A normal process has standard deviation 8 units and an adjustable mean. Using $\Phi(0.8)\approx0.7881$ , estimate all mean settings for which at most $21.19\%$ of readings fall below 21 units and at most $21.19\%$ exceed 45 units. Give the boundary estimates to the nearest tenth.
1. Hint 1
  Each tail limit places a restriction on the mean.
2. Hint 2
  Translate each limit into a z-score inequality and combine the restrictions.
Answer

Approximately $27.4\le\mu\le38.6$ units.

Full solution

By the table and symmetry, a tail of about $0.2119$ lies beyond $z=0.8$ on either side.
At most $21.19\%$ below 21 requires, approximately, $\frac{21-\mu}8\le-0.8$ , so
$\mu\ge27.4.$

At most $21.19\%$ above 45 requires, approximately, $\frac{45-\mu}8\ge0.8$ , so
$\mu\le38.6.$
Both estimated restrictions hold together for about $27.4\le\mu\le38.6$ units.

Answer

Approximately $27.4\le\mu\le38.6$ units.

Key idea

Limits on both tails can confine an adjustable mean to an interval of settings.
Problem 5: Mark complete
Problem 6 A central measurement band

A normal distribution has a central interval from 12 to 28 units extending two standard deviations on each side of its mean. Under the empirical rule, about what percentage of values fall below 8 units? State the mean and standard deviation used.
1. Hint 1
  The midpoint of the central interval is the mean, and its half-width equals two standard deviations.
2. Hint 2
  Locate 8 relative to that mean, then split the appropriate outer area equally between the two tails.
Answer

$\mu=20$ units, $\sigma=4$ units; approximately $0.15\%$ below 8 units.

Full solution

The mean is $(12+28)/2=20$ , and the half-width is $8$ , so $\sigma=4$ .
The value $8$ is three standard deviations below the mean because
$\frac{8-20}{4}=-3.$
The rounded rule leaves about $0.3\%$ outside the central three-standard-deviation interval.
Symmetry assigns half to the lower tail, about $0.15\%$ .

Answer

$\mu=20$ units, $\sigma=4$ units; approximately $0.15\%$ below 8 units.

Key idea

Recovering a normal scale from a central interval lets the empirical rule locate another tail.
Problem 6: Mark complete
Problem 7 A whole-number cutoff

A normal reading has mean 27 units and standard deviation 10 units. A device permits only whole-number upper cutoffs. Find the smallest permitted cutoff $c$ for which at least $99.6\%$ of readings fall below $c$ . Use $\Phi(2.6)\approx0.9953$ and $\Phi(2.7)\approx0.9965$ .
1. Hint 1
  Locate the target cumulative area between the supplied entries.
2. Hint 2
  Convert both bounding z-scores to raw values and use the whole-number restriction.
Answer

$c=54$ units.

Full solution

The target $0.996$ lies between $0.9953$ and $0.9965$ , so the required z is between $2.6$ and $2.7$ .
Walking back with $x=\mu+z\sigma$ gives the raw values $27+26=53$ and $27+27=54$ .

At $c=53$ only about $99.53\%$ of readings fall below, which is not enough; at $c=54$ about $99.65\%$ do.
No whole number lies between them, so the smallest permitted cutoff is $54$ units.

Answer

$c=54$ units.

Key idea

For a normal model, a cumulative percentage walks back to a raw cutoff through $x=\mu+z\sigma$ .
Problem 7: Mark complete
Problem 8 A rounded reading

A continuous normal quantity X has mean 10 and standard deviation 1. A display rounds X to the nearest whole number. A student says the display shows 10 with probability zero since $P(X=10)=0$ . Is this correct? Find the display probability to four decimal places using $\Phi(0.5)\approx0.6915$ .
1. Hint 1
  A displayed integer represents an interval of underlying values.
2. Hint 2
  Find the raw values that round to 10, then convert their boundaries to z-scores.
Answer

No; the probability is approximately $0.3830$ .

Full solution

The display shows $10$ when X falls from $9.5$ to $10.5$ , with any tie convention affecting only endpoints of probability zero.
These bounds give $z=-0.5$ and $z=0.5$ .
Therefore
$\begin{aligned}P&\approx0.6915-0.3085 \\ P&\approx0.3830.\end{aligned}$
A single exact value has probability zero, but the display groups an interval with positive width.

Answer

No; the probability is approximately $0.3830$ .

Key idea

Rounded measurements describe intervals, so a displayed value can have positive probability in a continuous model.
Problem 8: Mark complete
Problem 9 A mirrored comparison

A normal quantity X has mean 50 and positive standard deviation. A student claims $P(X<43)=P(X>57)$ without knowing the standard deviation. Is the claim correct? Explain why the tails have this relationship.
1. Hint 1
  Both cutoffs are the same raw distance from the mean.
2. Hint 2
  Compute each cutoff's signed distance from the mean, then use the symmetry of a normal density.
Answer

Yes; the tails are mirror images about 50, with both cutoffs 7 units from the mean.

Full solution

The deviations are $43-50=-7$ and $57-50=7$ .
Their z-scores are $-7/\sigma$ and $7/\sigma$ .
A normal density is symmetric about its mean, so reflection exchanges the lower and upper tails and gives
$P(X<43)=P(X>57).$
Their numerical size depends on the positive standard deviation, but their equality does not.

Answer

Yes; the tails are mirror images about 50, with both cutoffs 7 units from the mean.

Key idea

Normal tails at equal distances on opposite sides of the mean have equal areas.
Problem 9: Mark complete
Problem 10 Two percentages

For a normal distribution, a student treats the empirical rule as exact and says the probability outside three standard deviations must equal $0.0030$ . Using $\Phi(3)\approx0.9987$ , assess this claim and give the table-based probability to four decimal places.
1. Hint 1
  Which regions remain outside the central three-standard-deviation band?
2. Hint 2
  Find one outer tail from the table and double it by symmetry.
Answer

The claim of exactness is false; the table gives approximately $0.0026$ .

Full solution

One tail has probability approximately
$1-0.9987=0.0013.$
Doubling gives
$\begin{aligned}P(|Z|>3)&\approx2(0.0013) \\ P(|Z|>3)&\approx0.0026.\end{aligned}$
The empirical rule instead rounds the central area to $99.7\%$ , leaving about $0.3\%$ .
Neither rounded value should be declared an exact identity, and their small difference is expected.

Answer

The claim of exactness is false; the table gives approximately $0.0026$ .

Key idea

The empirical rule and a normal table have different rounding precision.
Problem 10: Mark complete

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