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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/maths/edexcel/a-level-statistics-3-the-normal-distribution/flashcards","label":"The Normal Distribution","subjectCode":"a_level_maths","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"d0b5569d-38f4-446b-955b-f7487bcb4971","title":"The Normal Distribution Flashcards","topicId":"b88e7835-a5ad-47c9-afa8-3551812acc17","topicSlug":"a-level-statistics-3-the-normal-distribution","cards":[{"id":"17a261f5-f66b-4bb9-8e1d-7754af587e4f","cardType":"basic","points":1,"front":"In the notation $X \\sim N(\\mu, \\sigma^2)$, what do the two parameters represent?","back":"

$\\mu$: the mean
$\\sigma^2$: the variance

","mcqOptions":[{"text":"

$\\mu$: the mean
$\\sigma^2$: the variance

","correct":true,"explanation":"In $X \\sim N(\\mu, \\sigma^2)$, $\\mu$ gives the centre of the distribution, while $\\sigma^2$ measures its spread as the variance."},{"text":"

$\\mu$: the variance
$\\sigma^2$: the mean

","correct":false,"explanation":"The parameters have been swapped: $\\mu$ always denotes the mean, while $\\sigma^2$ denotes the variance. Confusing variance with standard deviation is a common notation error."},{"text":"

$\\mu$: the standard deviation
$\\sigma^2$: the mean

","correct":false,"explanation":"The mean is represented by $\\mu$, not $\\sigma^2$. The symbol $\\sigma$ represents standard deviation, so $\\sigma^2$ represents variance."},{"text":"

$\\mu$: the median
$\\sigma^2$: the standard deviation

","correct":false,"explanation":"For a normal distribution, $\\mu$ represents the mean, which is also the median, but it is not labelled the median here. The squared parameter $\\sigma^2$ is variance, not standard deviation."}]},{"id":"83422ae2-b4f3-48ed-a341-c81bfb5bffd1","cardType":"basic","points":2,"front":"How does an increase in the standard deviation $\\sigma$ affect the shape of the normal curve?","back":"The curve becomes wider and flatter.","mcqOptions":[{"text":"The curve becomes wider and flatter.","correct":true,"explanation":"A larger standard deviation spreads values further from the mean. To keep the total area equal to 1, the normal curve becomes wider and its peak becomes flatter."},{"text":"The curve becomes narrower and taller.","correct":false,"explanation":"This is the effect of decreasing, not increasing, $\\sigma$. A larger standard deviation spreads the distribution out, making it wider and flatter while preserving total area."},{"text":"The curve shifts to the right and becomes flatter.","correct":false,"explanation":"Changing $\\sigma$ alters the spread but not the centre. The mean controls horizontal position; increasing $\\sigma$ makes the curve wider and flatter without shifting it."},{"text":"The curve becomes wider and taller.","correct":false,"explanation":"Although increasing $\\sigma$ makes the curve wider, its maximum height must decrease because the total area remains 1. Therefore it becomes wider and flatter."}]},{"id":"9ead7fd0-4eb8-4c16-b684-7d1dbad141a2","cardType":"cloze","points":null,"front":"For any continuous normal variable, the probability of the variable taking exactly one specific value is [ ].","back":"For any continuous normal variable, the probability of the variable taking exactly one specific value is 0.","mcqOptions":[{"text":"0","correct":true,"explanation":"A normal variable is continuous, so the probability at one exact point has zero area under the curve. Probabilities come from intervals, not individual values."},{"text":"1","correct":false,"explanation":"A probability of 1 would mean the variable is certain to equal that one value. For a continuous normal variable, one point has zero width and therefore probability 0."},{"text":"0.5","correct":false,"explanation":"The value 0.5 might describe an interval probability, but not the probability of one exact point in a continuous distribution. The correct point probability is 0."},{"text":"$$\\frac{1}{2}$","correct":false,"explanation":"An exact value has no interval width, so its probability is zero, not $\\frac{1}{2}$. A non-zero probability applies to a range of values."}]},{"id":"db596217-9695-4742-b966-b90b2380bd9d","cardType":"basic","points":1,"front":"Why is $P(X < a)$ equal to $P(X \\le a)$ in a normal distribution?","back":"The probability of $X$ being exactly $a$ is zero.","mcqOptions":[{"text":"The probability of $X$ being exactly $a$ is zero.","correct":true,"explanation":"A normal distribution is continuous, so $P(X=a)=0$. Consequently, including or excluding the single endpoint does not change the probability, giving $P(Xless than $a$ is zero.","correct":false,"explanation":"The probability below $a$ is generally a positive area under the curve. Equality holds because the probability of the single endpoint $X=a$ is zero."},{"text":"The probability of $X$ being greater than $a$ is zero.","correct":false,"explanation":"Values greater than $a$ usually have positive probability in a normal distribution. The relevant fact is that the probability of exactly $a$ is zero."},{"text":"The probability of $X$ being exactly the mean is one.","correct":false,"explanation":"Even the mean is only one point on a continuous distribution, so its probability is zero, not one. The endpoint equality follows from $P(X=a)=0$ for any $a$."}]},{"id":"80e6e9b4-47ce-46b0-89b6-2807e819ea2f","cardType":"basic","points":1,"front":"Identify the mean and standard deviation of $X \\sim N(100, 25)$.","back":"

Mean: 100
Standard deviation: 5 ($\\sqrt{25}$)

","mcqOptions":[{"text":"

Mean: 100
Standard deviation: 5 ($\\sqrt{25}$)

","correct":true,"explanation":"In $N(100,25)$, the first parameter is the mean and the second is the variance. The standard deviation is $\\sqrt{25}=5$."},{"text":"

Mean: 100
Standard deviation: 25 ($\\sqrt{625}$)

","correct":false,"explanation":"The second parameter, 25, is the variance rather than the standard deviation. Taking its square root gives the standard deviation, $\\sqrt{25}=5$."},{"text":"

Mean: 25
Standard deviation: 10 ($\\sqrt{100}$)

","correct":false,"explanation":"The parameters have been reversed and the spread has then been calculated from the wrong value. The mean is 100, and $\\sqrt{25}=5$ is the standard deviation."},{"text":"

Mean: 100
Standard deviation: 625 ($25^2$)

","correct":false,"explanation":"Squaring the variance increases the spread instead of converting it to standard deviation. The required operation is $\\sqrt{25}$, giving a standard deviation of 5."}]},{"id":"19af8f51-d0b5-4e95-bb91-9af4eadda6c0","cardType":"cloze","points":null,"front":"The standard normal distribution is denoted by $\\text{[...]}$.","back":"The standard normal distribution is denoted by $\\boxed{Z \\sim N(0, 1)}$.","mcqOptions":[{"text":"$$Z \\sim N(0, 1)$","correct":true,"explanation":"The standard normal distribution has mean 0 and variance 1, so it is written as $Z \\sim N(0,1)$. Its standard deviation is also 1."},{"text":"$$Z \\sim N(1, 0)$","correct":false,"explanation":"The standard normal has mean 0 and variance 1, not mean 1 and variance 0. A variance of 0 would describe a constant rather than a normal curve."},{"text":"$$Z \\sim N(0, 0)$","correct":false,"explanation":"The mean is correctly shown as 0, but the variance must be 1. Variance 0 gives no spread, so it cannot represent the standard normal distribution."},{"text":"$$Z \\sim N(1, 1)$","correct":false,"explanation":"Although the variance is correctly 1, the standard normal mean is 0, not 1. The notation must therefore be $Z \\sim N(0,1)$."}]},{"id":"d13f8a3c-6c0b-4165-b223-6c1eda32fffe","cardType":"cloze","points":null,"front":"The formula to standardise a value $X$ is $\\text{[...]}$.","back":"The formula to standardise a value $X$ is $\\boxed{Z = \\frac{X - \\mu}{\\sigma}}$.","mcqOptions":[{"text":"$$Z = \\frac{X - \\mu}{\\sigma}$","correct":true,"explanation":"Subtracting $\\mu$ centres the value relative to the mean, and dividing by $\\sigma$ measures that distance in standard deviations. The result is the standardised value $Z$."},{"text":"$$Z = \\frac{X + \\mu}{\\sigma}$","correct":false,"explanation":"Adding the mean does not centre $X$; it moves the value further from the reference point. Standardisation requires subtracting $\\mu$ before dividing by $\\sigma$."},{"text":"$$Z = \\frac{X - \\sigma}{\\mu}$","correct":false,"explanation":"The mean must be subtracted to centre the observation, and the standard deviation must divide to measure standard-deviation units. These roles have been incorrectly exchanged."},{"text":"$$Z = \\frac{X}{\\mu - \\sigma}$","correct":false,"explanation":"Standardisation is based on the distance from the mean, $X-\\mu$, scaled by the standard deviation $\\sigma$. Combining the parameters as $\\mu-\\sigma$ does not produce a z-value."}]},{"id":"b85b878e-0391-4702-9fe2-19ef2cf8e4fa","cardType":"basic","points":1,"front":"What does a $z$-value represent in the context of the normal distribution?","back":"The number of standard deviations a value is from the mean.","mcqOptions":[{"text":"The number of standard deviations a value is from the mean.","correct":true,"explanation":"A z-value measures the signed distance from the mean in standard-deviation units. Positive values lie above the mean and negative values lie below it."},{"text":"The number of variances a value is from the mean.","correct":false,"explanation":"A z-value uses standard deviation, not variance, as the unit of spread. Dividing by $\\sigma^2$ would give the wrong scale and would not represent standard-deviation units."},{"text":"The probability of a value being from the mean.","correct":false,"explanation":"A z-value is a standardised distance, not a probability. Probabilities are found from areas under the normal curve after using the z-value."},{"text":"The number of units a value is above the mean.","correct":false,"explanation":"A z-value can be positive or negative and measures distance in standard deviations, not ordinary units or only upward distance. Values below the mean have negative z-values."}]},{"id":"fb516f44-7ba1-4575-8b99-589d6e4a1a81","cardType":"basic","points":1,"front":"State the symmetry property of $P(X < \\mu)$ for any normal distribution.","back":"$$P(X < \\mu) = 0.5$","mcqOptions":[{"text":"$$P(X < \\mu) = 0.5$","correct":true,"explanation":"A normal distribution is symmetric about its mean, so exactly half of the probability lies below $\\mu$ and half lies above it."},{"text":"$$P(X < \\mu) = 0.25$","correct":false,"explanation":"This confuses the mean with a quartile. Symmetry divides the total probability equally at the mean, so $P(X < \\mu) = 0.5$."},{"text":"$$P(X < \\mu) = 0$","correct":false,"explanation":"The mean is not the distribution’s lower boundary. Half of the probability lies below the mean, giving $P(X < \\mu) = 0.5$."},{"text":"$$P(X < \\mu) = 1$","correct":false,"explanation":"The mean is not an upper boundary containing the entire distribution. Symmetry places half the probability below it, so $P(X < \\mu) = 0.5$."}]},{"id":"707e0ae6-42ba-40b0-a088-240c79eb227c","cardType":"cloze","points":null,"front":"To find a value $x$ from a standard normal $z$-value, use the formula $\\text{[...]}$.","back":"To find a value $x$ from a standard normal $z$-value, use the formula $\\boxed{x = \\mu + \\sigma z}$.","mcqOptions":[{"text":"$$x = \\mu + \\sigma z$","correct":true,"explanation":"Rearranging $z = \\frac{x-\\mu}{\\sigma}$ gives $x-\\mu=\\sigma z$, and then adding $\\mu$ gives $x=\\mu+\\sigma z$."},{"text":"$$x = \\frac{z-\\mu}{\\sigma}$","correct":false,"explanation":"This incorrectly applies the standardisation formula to $z$ rather than rearranging it. Solving $z=\\frac{x-\\mu}{\\sigma}$ for $x$ gives $x=\\mu+\\sigma z$."},{"text":"$$x = \\mu - \\sigma z$","correct":false,"explanation":"The sign has been reversed when rearranging the standardisation formula. Since $x-\\mu=\\sigma z$, the correct result is $x=\\mu+\\sigma z$."},{"text":"$$x = \\mu + \\frac{z}{\\sigma}$","correct":false,"explanation":"This divides the standard normal value by the standard deviation instead of multiplying it. Rescaling requires $\\sigma z$, so $x=\\mu+\\sigma z$."}]},{"id":"c7b7444d-e5d8-4cbe-8c52-e599805480bc","cardType":"basic","points":2,"front":"What is the first step required to find a missing $\\mu$ or $\\sigma$ when given a probability?","back":"Use inverse normal to find the corresponding $z$-value.","mcqOptions":[{"text":"Use inverse normal to find the corresponding $z$-value.","correct":true,"explanation":"A cumulative probability identifies a percentile of the standard normal distribution. Applying inverse normal converts that probability into the corresponding standardised $z$-value."},{"text":"Use normal cumulative distribution to find the corresponding $z$-value.","correct":false,"explanation":"The cumulative distribution function finds a probability from a known value, reversing the required direction. Use inverse normal to obtain the $z$-value from the probability."},{"text":"Use linear interpolation to find the corresponding $z$-value.","correct":false,"explanation":"Linear interpolation is not the initial method for converting a normal probability into a standard score. Use inverse normal to find the corresponding $z$-value."},{"text":"Use standardisation to find the corresponding $z$-value.","correct":false,"explanation":"Standardisation requires the mean and standard deviation already to be known, so it cannot initially resolve a missing parameter. First use inverse normal to find the $z$-value."}]},{"id":"9ec40bbf-ee24-4ae0-8a66-7c25a750fe4e","cardType":"basic","points":1,"front":"How many probability statements are needed to calculate both a missing mean and a missing standard deviation?","back":"Two (to form two simultaneous equations).","mcqOptions":[{"text":"Two (to form two simultaneous equations).","correct":true,"explanation":"The unknown mean and standard deviation are two separate parameters. Each probability statement provides one equation, so two statements produce two simultaneous equations."},{"text":"One (to form one simultaneous equation).","correct":false,"explanation":"One probability statement supplies only one equation, which cannot uniquely determine two unknown parameters. Two statements are needed to form two simultaneous equations."},{"text":"Three (to form three simultaneous equations).","correct":false,"explanation":"A third independent statement is unnecessary because there are only two unknowns, $\\mu$ and $\\sigma$. Two probability statements provide the required simultaneous equations."},{"text":"Four (to form four simultaneous equations).","correct":false,"explanation":"This incorrectly assumes each parameter needs two separate conditions. There are two unknown parameters, so two probability statements are sufficient to form two simultaneous equations."}]},{"id":"980a5eee-147a-4b6a-8d77-46a8d56b3ca1","cardType":"basic","points":1,"front":"In race times, how should the phrase \"faster than 10 seconds\" be written in probability notation?","back":"$$P(T < 10)$","mcqOptions":[{"text":"$$P(T < 10)$","correct":true,"explanation":"A faster race time is numerically smaller, so finishing faster than 10 seconds means the recorded time $T$ is less than 10."},{"text":"$$P(T > 10)$","correct":false,"explanation":"This treats a larger time as faster, reversing the relationship. Faster performances take fewer seconds, so the correct notation is $P(T < 10)$."},{"text":"$$P(T \\leq 10)$","correct":false,"explanation":"The word “faster” indicates a strict inequality and excludes a time of exactly 10 seconds. The correct notation is therefore $P(T < 10)$."},{"text":"$$P(T \\geq 10)$","correct":false,"explanation":"This both reverses the direction and includes exactly 10 seconds. Faster times are below 10 seconds, so the correct expression is $P(T < 10)$."}]},{"id":"10e43ed3-e58a-4eca-a9b4-4ac4c197e70a","cardType":"cloze","points":null,"front":"The general formula for conditional probability is $P(A|B) = \\frac{\\text{[...]}}{P(B)}$.","back":"The general formula for conditional probability is $P(A|B) = \\frac{\\boxed{P(A \\cap B)}}{P(B)}$.","mcqOptions":[{"text":"$$P(A \\cap B)$","correct":true,"explanation":"Conditional probability restricts attention to outcomes in $B$. The favourable outcomes are those in both $A$ and $B$, so the numerator is $P(A \\cap B)$."},{"text":"$$P(A \\cup B)$","correct":false,"explanation":"The union includes outcomes occurring in either event, including those outside the condition $B$. The numerator must be the intersection $P(A \\cap B)$."},{"text":"$$P(A)$","correct":false,"explanation":"Using all outcomes in $A$ ignores the requirement that $B$ has occurred. The numerator must count outcomes shared by both events, namely $P(A \\cap B)$."},{"text":"$$P(B|A)$","correct":false,"explanation":"This is a conditional probability in the reverse direction, not the joint probability required in the numerator. The correct numerator is $P(A \\cap B)$."}]},{"id":"f2bc8067-554c-4259-9074-873f8bad06a1","cardType":"cloze","points":null,"front":"The general formula for conditional probability is $P(A|B) = \\frac{P(A \\cap B)}{\\text{[...]}}$.","back":"The general formula for conditional probability is $P(A|B) = \\frac{P(A \\cap B)}{\\boxed{P(B)}}$.","mcqOptions":[{"text":"$$P(B)$","correct":true,"explanation":"Conditioning on $B$ restricts the sample space to event $B$, so the joint probability $P(A \\cap B)$ must be divided by $P(B)$."},{"text":"$$P(A)$","correct":false,"explanation":"Dividing by $P(A)$ would condition on $A$ and produce $P(B|A)$ instead. Since the stated condition is $B$, the denominator is $P(B)$."},{"text":"$$P(A \\cap B)$","correct":false,"explanation":"Using the intersection as both numerator and denominator would always produce 1 when it is non-zero. Conditional probability instead divides by the conditioning event’s probability, $P(B)$."},{"text":"$$1-P(B)$","correct":false,"explanation":"This divides by the probability that $B$ does not occur, which contradicts the condition that $B$ has occurred. The correct denominator is $P(B)$."}]},{"id":"4beb14f3-5108-4354-be97-d8d3494f6186","cardType":"basic","points":2,"front":"If a battery has already lasted 40 hours, how is the probability that it lasts at least another 10 hours expressed?","back":"$$P(X > 50 | X > 40)$","mcqOptions":[{"text":"$$P(X > 50 | X > 40)$","correct":true,"explanation":"Lasting another 10 hours after already exceeding 40 hours means the total lifetime must exceed 50 hours, conditional on it having exceeded 40 hours."},{"text":"$$P(X > 10 | X > 40)$","correct":false,"explanation":"This treats “another 10 hours” as a total lifetime of 10 hours. The threshold must be added to 40, giving $P(X > 50 | X > 40)$."},{"text":"$$P(X > 50)$","correct":false,"explanation":"This ignores the information that the battery has already lasted 40 hours. That information must be included as a condition, giving $P(X > 50 | X > 40)$."},{"text":"$$P(X > 40 | X > 50)$","correct":false,"explanation":"The condition and target events have been reversed. The known event is survival beyond 40 hours, while the target is survival beyond 50 hours."}]},{"id":"09e0b02d-bea1-41a0-b3c1-c83e71142891","cardType":"cloze","points":null,"front":"For a population $X \\sim N(\\mu, \\sigma^2)$, the distribution of the sample mean is $\\text{[...]}$.","back":"For a population $X \\sim N(\\mu, \\sigma^2)$, the distribution of the sample mean is $\\boxed{\\bar{X} \\sim N(\\mu, \\frac{\\sigma^2}{n})}$.","mcqOptions":[{"text":"$$\\bar{X} \\sim N(\\mu, \\frac{\\sigma^2}{n})$","correct":true,"explanation":"Averaging $n$ independent observations preserves the population mean $mu$ and divides the variance by $n$, so the sample mean has variance $\\frac{\\sigma^2}{n}$."},{"text":"$$\\bar{X} \\sim N(\\mu, \\frac{\\sigma^2}{\\sqrt{n}})$","correct":false,"explanation":"Dividing the variance by $\\sqrt{n}$ confuses the variance with the standard deviation. The sample mean has variance $\\frac{\\sigma^2}{n}$, giving $\\bar{X} \\sim N(\\mu, \\frac{\\sigma^2}{n})$."},{"text":"$$\\bar{X} \\sim N(\\mu, n\\sigma^2)$","correct":false,"explanation":"Multiplying by $n$ incorrectly makes sample means more variable as the sample grows. Averaging reduces variance, so the correct distribution is $\\bar{X} \\sim N(\\mu, \\frac{\\sigma^2}{n})$."},{"text":"$$\\bar{X} \\sim N(n\\mu, \\frac{\\sigma^2}{n})$","correct":false,"explanation":"The mean of $\\bar{X}$ remains $\\mu$ rather than becoming $n\\mu$; only a sample total has mean $n\\mu$. Its distribution is $N(\\mu, \\frac{\\sigma^2}{n})$."}]},{"id":"f4c9b97e-1061-41f2-ae84-d23de5bce340","cardType":"basic","points":1,"front":"Define the \"standard error\" of the sample mean.","back":"The standard deviation of the sampling distribution: $\\frac{\\sigma}{\\sqrt{n}}$.","mcqOptions":[{"text":"The standard deviation of the sampling distribution: $\\frac{\\sigma}{\\sqrt{n}}$.","correct":true,"explanation":"The standard error measures the standard deviation of sample means across repeated samples. Since $\\operatorname{Var}(\\bar{X})=\\frac{\\sigma^2}{n}$, its standard deviation is $\\frac{\\sigma}{\\sqrt{n}}$."},{"text":"The standard deviation of the population distribution: $\\sigma$.","correct":false,"explanation":"The quantity $\\sigma$ is the standard deviation of individual population values, not of sample means. Averaging $n$ observations reduces the standard deviation to $\\frac{\\sigma}{\\sqrt{n}}$."},{"text":"The variance of the sampling distribution: $\\frac{\\sigma^2}{n}$.","correct":false,"explanation":"The quantity $\\frac{\\sigma^2}{n}$ is the variance of the sampling distribution, not its standard deviation. Taking its square root gives the standard error $\\frac{\\sigma}{\\sqrt{n}}$."},{"text":"The standard deviation of the sampling distribution: $\\frac{\\sigma}{n}$.","correct":false,"explanation":"Dividing the standard deviation by $n$ rather than $\\sqrt{n}$ overstates the effect of averaging. The variance divides by $n$, so the standard error is $\\frac{\\sigma}{\\sqrt{n}}$."}]},{"id":"8cc975c9-922d-4a12-b8fd-397822331627","cardType":"basic","points":1,"front":"State the null and alternative hypotheses for a test where the mean $\\mu$ is suspected to have increased from a value $k$.","back":"

$H_0: \\mu = k$
$H_1: \\mu > k$

","mcqOptions":[{"text":"

$H_0: \\mu = k$
$H_1: \\mu > k$

","correct":true,"explanation":"The null hypothesis fixes the mean at its original value $k$. Because the suspected change is specifically an increase, the alternative hypothesis is the one-tailed statement $\\mu>k$."},{"text":"

$H_0: \\mu = k$
$H_1: \\mu < k$

","correct":false,"explanation":"The alternative $\\muk$."},{"text":"

$H_0: \\mu = k$
$H_1: \\mu \\ne k$

","correct":false,"explanation":"The alternative $\\mu\\ne k$ describes a two-tailed test for any change, not specifically an increase. The correct directional alternative is $H_1: \\mu>k$, with $H_0: \\mu=k$."},{"text":"

$H_0: \\mu > k$
$H_1: \\mu = k$

","correct":false,"explanation":"Equality belongs in the null hypothesis, which represents the original claimed value. The correct hypotheses are $H_0: \\mu=k$ and $H_1: \\mu>k$ for a suspected increase."}]},{"id":"aa583538-a0af-4e70-87c5-0828bf2cdbc0","cardType":"basic","points":1,"front":"When conducting a hypothesis test, what is the condition to reject the null hypothesis $H_0$ using a p-value?","back":"Reject $H_0$ if p-value < significance level.","mcqOptions":[{"text":"Reject $H_0$ if p-value < significance level.","correct":true,"explanation":"A small p-value indicates that the observed result would be unlikely if $H_0$ were true. The null hypothesis is rejected when the p-value falls below the chosen significance level."},{"text":"Reject $H_0$ if p-value > significance level.","correct":false,"explanation":"A p-value greater than the significance level does not provide sufficient evidence against $H_0$. The correct decision rule is to reject $H_0$ when the p-value is below the significance level."},{"text":"Fail to reject $H_0$ if p-value < significance level.","correct":false,"explanation":"The inequality has been reversed: a p-value below the significance level leads to rejection, whereas a larger p-value means there is insufficient evidence to reject $H_0$."},{"text":"Reject $H_0$ if p-value = significance level.","correct":false,"explanation":"Equality is not included in the stated strict decision rule. At the significance-level boundary, the usual rule here does not reject $H_0$; rejection requires a smaller p-value."}]},{"id":"1dcde702-9ad1-47c1-b0ca-27b8d0c3f9c3","cardType":"basic","points":1,"front":"How do you estimate the population standard deviation for a large sample if $\\sigma$ is unknown?","back":"Use the sample standard deviation ($s$).","mcqOptions":[{"text":"Use the sample standard deviation ($s$).","correct":true,"explanation":"For a large sample, the sample standard deviation $s$ provides a suitable estimate of the unknown population standard deviation $\\sigma$, allowing it to be substituted in standard-error calculations."},{"text":"Use the sample variance ($s^2$).","correct":false,"explanation":"The sample variance $s^2$ estimates the population variance $\\sigma^2$, not the population standard deviation $\\sigma$. The required estimate of standard deviation is $s$."},{"text":"Use the sample standard error ($\\frac{s}{\\sqrt{n}}$).","correct":false,"explanation":"The standard error $\\frac{s}{\\sqrt{n}}$ estimates the spread of sample means, not the spread of individual population values. The population standard deviation is estimated directly by $s$."},{"text":"Use the sample range ($R$).","correct":false,"explanation":"The range depends only on the two extreme observations and is not a direct estimate of population standard deviation. For a large sample, use the sample standard deviation $s$."}]}]}],"examCountdown":{"examDate":"2027-05-26","paperLabel":"Paper 1"},"heading":"The Normal Distribution","title":"The Normal Distribution Flashcards","topbar":{"backHref":"/a-level/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"b88e7835-a5ad-47c9-afa8-3551812acc17","walled":"$L39"}],"$L3a","$L3b"]