flatter curve."},{"id":"block-3","content":"For example, if $W \\sim N(120,16)$, then $\\mu=120$ and $\\sigma=\\sqrt{16}=4$. By symmetry, $P(W<120)=0.5$."}]},{"id":"card-2","type":"mcq","order":2,"options":[{"id":"a","text":"Mean 20 and standard deviation 5","isCorrect":true},{"id":"b","text":"Mean 25 and standard deviation 20","isCorrect":false},{"id":"c","text":"Mean 20 and standard deviation 25","isCorrect":false},{"id":"d","text":"Mean 5 and variance 20","isCorrect":false}],"question":"If $X \\sim N(20,25)$, which statement is correct?","explanation":"In $N(\\mu,\\sigma^2)$, the first number is the mean and the second number is the variance. So $\\mu=20$ and $\\sigma=\\sqrt{25}=5$. The most common wrong answer is C, which treats the variance as if it were the standard deviation.","correctOptionId":"a"},{"id":"card-3","type":"content","order":3,"title":"Standardising to the standard normal","blocks":[{"id":"block-1","content":"The standard normal distribution is $Z \\sim N(0,1)$. To convert a value of $X$ to a z-score, use $Z=\\frac{X-\\mu}{\\sigma}$, which tells you how many standard deviations the value is from the mean."},{"id":"block-2","content":"Once you standardise, normal probabilities become standard normal probabilities. For a continuous variable, $P(X257)=P(Z>1.4)$, and values below the mean give negative z-scores."}]},{"id":"card-4","type":"fill_blank","order":4,"blanks":[{"id":"z","isMath":true,"correctAnswer":"2"}],"sentence":"For $X \\sim N(50,9)$, the value $x=56$ standardises to $z =$ {{blank:z}}."},{"id":"card-5","type":"content","order":5,"title":"Worked example: finding probabilities","blocks":[{"id":"block-1","content":"![Normal distribution for cereal box mass with 243, 246, 250, 255 and 257 marked, the right tail beyond 257 shaded, and the interval 246 to 255 shaded](https://assets.mathsgenie.co.uk/lesson-v2-img/inline/1d03ae9b-a5f6-47b2-a6e7-a392baad758a.jpeg)\nLet $X \\sim N(250,5^2)$ represent the mass of a cereal box in grams. A good first step is to mark the mean 250 and decide which region of the curve you need."},{"id":"block-2","content":"For the right tail beyond 257 g, standardise the boundary:\n$$\nz=\\frac{257-250}{5}=1.4\n$$\nSo $P(X>257)=P(Z>1.4)\\approx 0.0808$."},{"id":"block-3","content":"For the left tail below 243 g, standardise again:\n$$\nz=\\frac{243-250}{5}=-1.4\n$$\nSo $P(X<243)=P(Z<-1.4)\\approx 0.0808$, which matches by symmetry."},{"id":"block-4","content":"For the middle region, standardise both ends:\n$$\n\\frac{246-250}{5}=-0.8,\\qquad \\frac{255-250}{5}=1\n$$\nHence $P(246-1.4)$","isCorrect":false},{"id":"d","text":"$$P(Z>1.4)$","isCorrect":false}],"question":"If $X \\sim N(250,5^2)$, which is the correct setup for $P(X<243)$?","explanation":"Since 243 is 7 below the mean, the z-score is $\\frac{243-250}{5}=-1.4$. The probability is also a lower-tail probability, so the correct setup is $P(Z<-1.4)$. The most common wrong answer is B, which forgets that values below the mean must have negative z-scores.","correctOptionId":"a"},{"id":"card-7","type":"flashcard","cards":[{"id":"fc-1","back":"$$X$ is normally distributed with mean $\\mu$ and variance $\\sigma^2$.","front":"$$X \\sim N(\\mu,\\sigma^2)$"},{"id":"fc-2","back":"$$Z \\sim N(0,1)$.","front":"Standard normal distribution"},{"id":"fc-3","back":"$$Z=\\frac{X-\\mu}{\\sigma}$.","front":"Standardising formula"},{"id":"fc-4","back":"A variable that can take any value in an interval, so probabilities are areas under a curve.","front":"Continuous random variable"},{"id":"fc-5","back":"A cut-off value below which a given proportion of observations lie.","front":"Percentile"}],"order":7},{"id":"card-8","type":"content","order":8,"title":"In

The Normal Distribution Lesson